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//Given an object, return true if all keys are strings in lower
// case or all keys are strings in upper case, else return false.
// The function should return false is the given object is empty.
// Examples:
// >>> check_dict_case({"a": "apple", "b": "banana"})
// true
// >>> check_dict_case({"a": "apple", "A": "banana", "B": "banana"})
// false
// >>> check_dict_case({"a": "apple", 8: "banana", "a": "apple"})
// false
// >>> check_dict_case({"Name": "John", "Age": "36", "City": "Houston"})
// false
// >>> check_dict_case({"STATE": "NC", "ZIP": "12345"})
// true
function check_dict_case(dict){
if (Object.keys(dict).length == 0){
return false;
}
var case_type = -1;
for (let key in dict){
if (typeof key == "string"){
if (case_type == -1){
if (key == key.toUpperCase()){
case_type = 1;
}
else if (key == key.toLowerCase()){
case_type = 0;
}
else{
return false;
}
}
else if (case_type == 1){
if (key != key.toUpperCase()){
return false;
}
}
else if (case_type == 0){
if (key != key.toLowerCase()){
return false;
}
}
}
else{
return false;
} | } | return true;
} |
//Given a string s, count the number of uppercase vowels in even indices.
// For example:
// >>> count_upper("aBCdEf")
// 1
// >>> count_upper("abcdefg")
// 0
// >>> count_upper("dBBE")
// 0
function count_upper(s){
var index;
var result = 0;
var vowel = ['A', 'E', 'I', 'O', 'U'];
for (index = 0; index < s.length; index += 2){
if (vowel.indexOf(s[index]) > -1){ | result += 1; | }
}
return result;
} |
//Given a string representing a space separated lowercase letters, return an object
// of the letter with the most repetition and containing the corresponding count.
// If several letters have the same occurrence, return all of them.
// Example:
// >>> histogram("a b c")
// {"a": 1, "b": 1, "c": 1}
// >>> histogram("a b b a")
// {"a": 2, "b": 2}
// >>> histogram("a b c a b")
// {"a": 2, "b": 2}
// >>> histogram("b b b b a")
// {"b": 4}
// >>> histogram("")
// {}
function histogram(test){
if (test.length === 0){
return {};
}
var lst = test.split(" ");
var count = {};
for (var i = 0; i < lst.length; i++){
if (count[lst[i]]){ | count[lst[i]] += 1; | }
else{
count[lst[i]] = 1;
}
}
console.log(count);
var max = 0;
var result = {};
for (var key in count){
if (count[key] > max){
max = count[key];
result = {};
result[key] = max;
}
else if (count[key] === max){
result[key] = max;
}
}
return result;
}
|
//Input to this function is a string represented multiple groups for nested parentheses separated by spaces.
// For each of the group, output the deepest level of nesting of parentheses.
// E.g. (()()) has maximum two levels of nesting while ((())) has three.
// >>> parse_nested_parens("(()()) ((())) () ((())()())")
// [2, 3, 1, 3]
function parse_nested_parens(paren_string){
var level = 0; | var maxlevel = 0; | var result = [];
for (var i = 0; i < paren_string.length; i++){
if (paren_string[i] == "("){
level += 1;
if (level > maxlevel){
maxlevel = level;
}
}
else if (paren_string[i] == ")"){
level -= 1;
}
else if (paren_string[i] == " "){
result.push(maxlevel);
maxlevel = 0;
level = 0;
}
}
result.push(maxlevel);
return result;
}
|
from typing import List
def minPath(grid: List[List[int]], k: int) -> List[int]:
"""
Given a grid with N rows and N columns (N >= 2) and a positive integer k,
each cell of the grid contains a value. Every integer in the range [1, N * N]
inclusive appears exactly once on the cells of the grid.
You have to find the minimum path of length k in the grid. You can start
from any cell, and in each step you can move to any of the neighbor cells,
in other words, you can go to cells which share an edge with you current
cell.
Please note that a path of length k means visiting exactly k cells (not
necessarily distinct).
You CANNOT go off the grid.
A path A (of length k) is considered less than a path B (of length k) if
after making the ordered lists of the values on the cells that A and B go
through (let's call them lst_A and lst_B), lst_A is lexicographically less
than lst_B, in other words, there exist an integer index i (1 <= i <= k)
such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
lst_A[j] = lst_B[j].
It is guaranteed that the answer is unique.
Return an ordered list of the values on the cells that the minimum path go through.
Examples:
>>> minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3)
[1, 2, 1]
>>> minPath([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1)
[1]
"""
min_val = float('inf')
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] < min_val:
min_val = grid[i][j]
row = i | col = j | path = [min_val]
while len(path) < k:
min_val = float('inf')
for i, j in (row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1):
if 0 <= i < len(grid) and 0 <= j < len(grid[0]) and grid[i][j] < min_val:
min_val = grid[i][j]
row = i
col = j
path.append(min_val)
return path |
//Input to this function is a string representing musical notes in a special ASCII format.
// Your task is to parse this string and return array of integers corresponding to how many beats does each
// not last.
// Here is a legend:
// 'o' - whole note, lasts four beats
// 'o|' - half note, lasts two beats
// '.|' - quater note, lasts one beat
// >>> parse_music("o o| .| o| o| .| .| .| .| o o")
// [4, 2, 1, 2, 2, 1, 1, 1, 1, 4, 4]
function parse_music(music_string){
let songs = music_string.split(' ');
let song = [];
songs.forEach(s => {
if(s === 'o'){
song.push(4);
}else if(s === 'o|'){
song.push(2);
}else if(s === '.|'){
song.push(1);
}
});
return song; | } | |
//Given an array of integers nums, find the minimum sum of any non-empty sub-array
// of nums.
// Example
// >>> minSubArraySum([2, 3, 4, 1, 2, 4])
// 1
// >>> minSubArraySum([-1, -2, -3])
// -6
function minSubArraySum(nums){
let sum = 0; | let minSum = nums[0]; | for(let i = 0; i < nums.length; i++){
sum += nums[i];
if(sum < minSum){
minSum = sum;
}
if(sum > 0){
sum = 0;
}
}
return minSum;
}
|
//Input to this function is a string represented multiple groups for nested parentheses separated by spaces.
// For each of the group, output the deepest level of nesting of parentheses.
// E.g. (()()) has maximum two levels of nesting while ((())) has three.
// >>> parse_nested_parens("(()()) ((())) () ((())()())")
// [2, 3, 1, 3]
function parse_nested_parens(paren_string){
var level = 0;
var maxlevel = 0;
var result = [];
for (var i = 0; i < paren_string.length; i++){
if (paren_string[i] == "("){
level += 1;
if (level > maxlevel){
maxlevel = level;
} | } | else if (paren_string[i] == ")"){
level -= 1;
}
else if (paren_string[i] == " "){
result.push(maxlevel);
maxlevel = 0;
level = 0;
}
}
result.push(maxlevel);
return result;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given array list of integers, return array list in strange order.
// Strange sorting, is when you start with the minimum value,
// then maximum of the remaining integers, then minimum and so on.
// Examples:
// >>> strangeSortList((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l))))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)4l, (long)2l, (long)3l)))
// >>> strangeSortList((new ArrayList<Long>(Arrays.asList((long)5l, (long)5l, (long)5l, (long)5l))))
// (new ArrayList<Long>(Arrays.asList((long)5l, (long)5l, (long)5l, (long)5l)))
// >>> strangeSortList((new ArrayList<Long>(Arrays.asList())))
// (new ArrayList<Long>(Arrays.asList()))
public static ArrayList<Long> strangeSortList(ArrayList<Long> lst) {
if (lst.size() <= 1) {
return lst; | } | ArrayList<Long> result = new ArrayList<Long>();
ArrayList<Long> sorted = new ArrayList<Long>(lst);
Collections.sort(sorted);
int min = 0;
int max = sorted.size() - 1;
while (min <= max) {
result.add(sorted.get(min));
min++;
if (min <= max) {
result.add(sorted.get(max));
max--;
}
}
return result;
}
} |
//Create a function that returns true if the last character
// of a given string is an alphabetical character and is not
// a part of a word, and false otherwise.
// Note: "word" is a group of characters separated by space.
// Examples:
// >>> check_if_last_char_is_a_letter("apple pie")
// false
// >>> check_if_last_char_is_a_letter("apple pi e")
// true
// >>> check_if_last_char_is_a_letter("apple pi e ")
// false
// >>> check_if_last_char_is_a_letter("")
// false
function check_if_last_char_is_a_letter(txt){
var result=txt.split(" ");
if(result[result.length-1].length==1 && result[result.length-1].match(/[a-z]/i)){
return true; | } | return false;
} |
//Given a positive integer n, return a sorted array that has the odd numbers in collatz sequence.
// The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
// as follows: start with any positive integer n. Then each term is obtained from the
// previous term as follows: if the previous term is even, the next term is one half of
// the previous term. If the previous term is odd, the next term is 3 times the previous
// term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.
// Note:
// 1. Collatz(1) is [1].
// 2. returned array sorted in increasing order.
// For example:
// get_odd_collatz(5) returns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
// >>> get_odd_collatz(5)
// [1, 5]
function get_odd_collatz(n){
var collatz_sequence = []
var odd_numbers = []
var odd_count = 0
collatz_sequence.push(n)
for(var i = 0; i < collatz_sequence.length; i++){
if(collatz_sequence[i]%2 == 0){
collatz_sequence.push(collatz_sequence[i]/2)
}
else if(collatz_sequence[i]%2 == 1){
collatz_sequence.push(collatz_sequence[i]*3+1)
}
if(collatz_sequence[i]%2 == 1){
odd_numbers.push(collatz_sequence[i])
odd_count += 1
}
if(collatz_sequence[i] == 1){ | break | }
}
if(odd_count != 0){
odd_numbers.sort(function(a, b){return a-b})
return odd_numbers
}
else{
odd_numbers.push(1)
return odd_numbers
}
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// You will be given a number in decimal form and your task is to convert it to
// binary format. The function should return a string, with each character representing a binary
// number. Each character in the string will be '0' or '1'.
// There will be an extra couple of characters 'db' at the beginning and at the end of the string.
// The extra characters are there to help with the format.
// Examples:
// >>> decimalToBinary((15l))
// ("db1111db")
// >>> decimalToBinary((32l))
// ("db100000db")
public static String decimalToBinary(long decimal) {
| return "db" + Long.toBinaryString(decimal) + "db"; | }
} |
//Given two arrays operator, and operand. The first array has basic algebra operations, and
// the second array is an array of integers. Use the two given arrays to build the algebric
// expression and return the evaluation of this expression.
// The basic algebra operations:
// Addition ( + )
// Subtraction ( - )
// Multiplication ( * )
// Floor division ( // )
// Exponentiation ( ** )
// Example:
// operator['+', '*', '-']
// array = [2, 3, 4, 5]
// result = 2 + 3 * 4 - 5
// => result = 9
// Note:
// The length of operator array is equal to the length of operand array minus one.
// Operand is an array of of non-negative integers.
// Operator array has at least one operator, and operand array has at least two operands.
function do_algebra(operator, operand){
let do_again;
let count = 0;
let temp;
let answer;
while(count < operator.length){
switch(operator[count]){
case '*':
operand[count] = operand[count] * operand[count + 1];
operand.splice(count + 1, 1);
operator.splice(count, 1);
count = 0;
do_again = true;
break;
case '//':
operand[count] = Math.floor(operand[count] / operand[count + 1]);
operand.splice(count + 1, 1);
operator.splice(count, 1);
count = 0;
do_again = true;
break;
case '**':
operand[count] = Math.pow(operand[count], operand[count + 1]);
operand.splice(count + 1, 1);
operator.splice(count, 1);
count = 0; | do_again = true; | break;
default:
count++;
do_again = false;
}
}
answer = operand[0];
operand.splice(0, 1);
while(operand.length > 0){
switch(operator[0]){
case '+':
answer += operand[0];
break;
case '-':
answer -= operand[0];
break;
}
operand.splice(0, 1);
operator.splice(0, 1);
}
return answer;
}
|
//Out of array of strings, return the longest one. Return the first one in case of multiple
// strings of the same length. Return undefined in case the input array is empty.
// >>> longest([])
// undefined
// >>> longest(["a", "b", "c"])
// "a"
// >>> longest(["a", "bb", "ccc"])
// "ccc"
function longest(strings){
var result;
if (strings.length === 0) {
return undefined;
}
for (var i = 0; i < strings.length; i++) {
if (!result || strings[i].length > result.length) {
result = strings[i]; | } | }
return result;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Return sorted unique common elements for two array lists.
// >>> common((new ArrayList<Long>(Arrays.asList((long)1l, (long)4l, (long)3l, (long)34l, (long)653l, (long)2l, (long)5l))), (new ArrayList<Long>(Arrays.asList((long)5l, (long)7l, (long)1l, (long)5l, (long)9l, (long)653l, (long)121l))))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)5l, (long)653l)))
// >>> common((new ArrayList<Long>(Arrays.asList((long)5l, (long)3l, (long)2l, (long)8l))), (new ArrayList<Long>(Arrays.asList((long)3l, (long)2l))))
// (new ArrayList<Long>(Arrays.asList((long)2l, (long)3l)))
public static ArrayList<Long> common(ArrayList<Long> l1, ArrayList<Long> l2) {
return new ArrayList<Long>(
l1.stream()
.filter(l2::contains)
.collect(Collectors.toSet()) | ); | }
} |
from typing import List, Tuple, Optional
def largest_smallest_integers(lst: List[int]) -> Tuple[Optional[int], Optional[int]]:
"""
Create a function that returns a tuple (a, b), where 'a' is
the largest of negative integers, and 'b' is the smallest
of positive integers in a list.
If there is no negative or positive integers, return them as None.
Examples:
>>> largest_smallest_integers([2, 4, 1, 3, 5, 7])
(None, 1)
>>> largest_smallest_integers([])
(None, None)
>>> largest_smallest_integers([0])
(None, None)
"""
| smallest_positive = None | largest_negative = None
for num in lst:
if num > 0:
if smallest_positive is None or num < smallest_positive:
smallest_positive = num
elif num < 0:
if largest_negative is None or num > largest_negative:
largest_negative = num
return (largest_negative, smallest_positive)
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Return 2^n modulo p (be aware of numerics).
// >>> modp((3l), (5l))
// (3l)
// >>> modp((1101l), (101l))
// (2l)
// >>> modp((0l), (101l))
// (1l)
// >>> modp((3l), (11l))
// (8l)
// >>> modp((100l), (101l))
// (1l)
public static long modp(long n, long p) {
if (n == 0) return 1;
long sqrt = modp(n / 2, p); | sqrt = (sqrt * sqrt) % p; | if (n % 2 == 0) {
return sqrt;
}
return (sqrt * 2) % p;
}
} |
def valid_date(date: str) -> bool:
"""You have to write a function which validates a given date string and
returns True if the date is valid otherwise False.
The date is valid if all of the following rules are satisfied:
1. The date string is not empty.
2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
3. The months should not be less than 1 or higher than 12.
4. The date should be in the format: mm-dd-yyyy
>>> valid_date('03-11-2000')
True
>>> valid_date('15-01-2012')
False
>>> valid_date('04-0-2040')
False
>>> valid_date('06-04-2020')
True
>>> valid_date('06/04/2020')
False
"""
if not date:
return False | days_per_month = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] | date_list = [int(d) for d in date.split('-')]
if len(date_list) != 3:
return False
if date_list[0] < 1 or date_list[0] > 12:
return False
if date_list[1] < 1 or date_list[1] > days_per_month[date_list[0] - 1]:
return False
if len(str(date_list[2])) != 4:
return False
return True |
from typing import Dict
def histogram(test: str) -> Dict[str, int]:
"""Given a string representing a space separated lowercase letters, return a dictionary
of the letter with the most repetition and containing the corresponding count.
If several letters have the same occurrence, return all of them.
Example:
>>> histogram('a b c')
{ 'a': 1, 'b': 1, 'c': 1 }
>>> histogram('a b b a')
{ 'a': 2, 'b': 2 }
>>> histogram('a b c a b')
{ 'a': 2, 'b': 2 }
>>> histogram('b b b b a')
{ 'b': 4 }
>>> histogram('')
{ }
"""
letters = {} | most_count = 0 | result = {}
for letter in test.split():
letters[letter] = letters.get(letter, 0) + 1
if letters[letter] >= most_count:
most_count = letters[letter]
for letter in letters:
if letters[letter] == most_count:
result[letter] = most_count
return result
|
from typing import List
def select_words(s: str, n: int) -> List[str]:
"""Given a string s and a natural number n, you have been tasked to implement
a function that returns a list of all words from string s that contain exactly
n consonants, in order these words appear in the string s.
If the string s is empty then the function should return an empty list.
Note: you may assume the input string contains only letters and spaces.
Examples:
>>> select_words('Mary had a little lamb', 4)
['little']
>>> select_words('Mary had a little lamb', 3)
['Mary', 'lamb']
>>> select_words('simple white space', 2)
[]
>>> select_words('Hello world', 4)
['world']
>>> select_words('Uncle sam', 3)
['Uncle']
"""
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'
words = s.split()
result = []
for word in words:
count = 0
for char in word:
if char.lower() in CONSONANTS:
count += 1
if count == n:
result.append(word) | return result | |
//Return array with elements incremented by 1.
// >>> incr_list([1, 2, 3])
// [2, 3, 4]
// >>> incr_list([5, 3, 5, 2, 3, 3, 9, 0, 123])
// [6, 4, 6, 3, 4, 4, 10, 1, 124]
function incr_list(l){
| var new_list = []; | for(var i = 0; i < l.length; i++){
new_list.push(l[i] + 1);
}
return new_list;
}
|
from typing import List
def sum_squares(lst: List[float]) -> int:
"""You are given a list of numbers.
You need to return the sum of squared numbers in the given list,
round each element in the list to the upper int(Ceiling) first.
Examples:
>>> lst([1.0, 2.0, 3.0])
14
>>> lst([1.0, 4.0, 9.0])
98
>>> lst([1.0, 3.0, 5.0, 7.0])
84
>>> lst([1.4, 4.2, 0.0])
29
>>> lst([-2.4, 1.0, 1.0])
6
"""
import math
for i in range(len(lst)):
lst[i] = math.ceil(lst[i]) | total = 0 | for i in lst:
total += i**2
return total |
//Check if two words have the same characters.
// >>> same_chars("eabcdzzzz", "dddzzzzzzzddeddabc")
// true
// >>> same_chars("abcd", "dddddddabc")
// true
// >>> same_chars("dddddddabc", "abcd")
// true
// >>> same_chars("eabcd", "dddddddabc")
// false
// >>> same_chars("abcd", "dddddddabce")
// false
// >>> same_chars("eabcdzzzz", "dddzzzzzzzddddabc")
// false
function same_chars(s0, s1){
let res = true;
for(let i = 0; i < s0.length; i++){
if(s1.indexOf(s0[i]) === -1){
res = false; | break; | }
}
for(let i = 0; i < s1.length; i++){
if(s0.indexOf(s1[i]) === -1){
res = false;
break;
}
}
return res;
}
|
//Return array with elements incremented by 1.
// >>> incr_list([1, 2, 3])
// [2, 3, 4]
// >>> incr_list([5, 3, 5, 2, 3, 3, 9, 0, 123])
// [6, 4, 6, 3, 4, 4, 10, 1, 124]
function incr_list(l){
var new_list = []; | for(var i = 0; i < l.length; i++){ | new_list.push(l[i] + 1);
}
return new_list;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// You're given an array array list of deposit and withdrawal operations on a bank account that starts with
// zero balance. Your task is to detect if at any point the balance of account fallls below zero, and
// at that point function should return true. Otherwise it should return false.
// >>> belowZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l))))
// (false)
// >>> belowZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)-4l, (long)5l))))
// (true)
public static boolean belowZero(ArrayList<Long> operations) {
long balance = 0;
for (long op : operations) {
balance += op;
if (balance < 0) {
return true;
}
}
return false;
} | } | |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given two array lists operator, and operand. The first array list has basic algebra operations, and
// the second array list is an array array list of integers. Use the two given array lists to build the algebric
// expression and return the evaluation of this expression.
// The basic algebra operations:
// Addition ( + )
// Subtraction ( - )
// Multiplication ( * )
// Floor division ( // )
// Exponentiation ( ** )
// Example:
// operator['+', '*', '-']
// array array list = [2, 3, 4, 5]
// result = 2 + 3 * 4 - 5
// => result = 9
// Note:
// The length of operator array list is equal to the length of operand array list minus one.
// Operand is an array array list of of non-negative integers.
// Operator array list has at least one operator, and operand array list has at least two operands.
public static long doAlgebra(ArrayList<String> op, ArrayList<Long> operand) {
| ArrayList<Long> list = new ArrayList<>(); | list.add(operand.get(0));
for (int i = 0; i < op.size(); i++) {
if (op.get(i).equals("+"))
list.add(operand.get(i + 1));
else if (op.get(i).equals("-"))
list.add(-operand.get(i + 1));
else if (op.get(i).equals("*"))
list.set(list.size() - 1, list.get(list.size() - 1) * operand.get(i + 1));
else if (op.get(i).equals("//"))
list.set(list.size() - 1, list.get(list.size() - 1) / operand.get(i + 1));
else if (op.get(i).equals("**"))
list.set(list.size() - 1, (long) Math.pow(list.get(list.size() - 1), operand.get(i + 1)));
}
return list.stream().mapToLong(x -> x).sum();
}
} |
//Given a string of words, return an array of words split on whitespace, if no whitespaces exists in the text you
// should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
// alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
// Examples
// >>> split_words("Hello world!")
// ["Hello", "world!"]
// >>> split_words("Hello,world!")
// ["Hello", "world!"]
// >>> split_words("abcdef")
// 3
function split_words(txt){
if(txt.includes(' ')){
return txt.split(' ');
}
else if(txt.includes(',')){
return txt.split(',');
}
else{
let count = 0;
for(let i = 0; i < txt.length; i++){
if(i % 2 != 0){
if(txt[i] == txt[i].toLowerCase()){
count++;
}
}
}
return count;
} | } | |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given an array array list of non-negative integers, return a cojava of the given array array list after sorting,
// you will sort the given array array list in ascending order if the sum( first index value, last index value) is odd,
// or sort it in descending order if the sum( first index value, last index value) is even.
// Note:
// * don't change the given array array list.
// Examples:
// >>> sortArray((new ArrayList<Long>(Arrays.asList())))
// (new ArrayList<Long>(Arrays.asList()))
// >>> sortArray((new ArrayList<Long>(Arrays.asList((long)5l))))
// (new ArrayList<Long>(Arrays.asList((long)5l)))
// >>> sortArray((new ArrayList<Long>(Arrays.asList((long)2l, (long)4l, (long)3l, (long)0l, (long)1l, (long)5l))))
// (new ArrayList<Long>(Arrays.asList((long)0l, (long)1l, (long)2l, (long)3l, (long)4l, (long)5l)))
// >>> sortArray((new ArrayList<Long>(Arrays.asList((long)2l, (long)4l, (long)3l, (long)0l, (long)1l, (long)5l, (long)6l))))
// (new ArrayList<Long>(Arrays.asList((long)6l, (long)5l, (long)4l, (long)3l, (long)2l, (long)1l, (long)0l)))
public static ArrayList<Long> sortArray(ArrayList<Long> array) {
if(array.size()==0) {
return array;
}
if(array.size()==1) {
return array;
}
if(array.size()>1) { | if((array.get(0)+array.get(array.size()-1))%2==0) { | Collections.sort(array, Collections.reverseOrder());
}
else {
Collections.sort(array);
}
}
return array;
}
} |
//You are given a non-empty array of positive integers. Return the greatest integer that is greater than
// zero, and has a frequency greater than or equal to the value of the integer itself.
// The frequency of an integer is the number of times it appears in the array.
// If no such a value exist, return -1.
// Examples:
// >>> search([4, 1, 2, 2, 3, 1])
// 2
// >>> search([1, 2, 2, 3, 3, 3, 4, 4, 4])
// 3
// >>> search([5, 5, 4, 4, 4])
// -1
function search(lst){
var h = {};
for (var i=0; i<lst.length; i++){
var x = lst[i];
if (x in h){
h[x] += 1;
}
else{
h[x] = 1;
}
}
var keys = Object.keys(h);
var max = 0;
for (var i=0; i<keys.length; i++){
if (keys[i] > max && h[keys[i]] >= keys[i]){
max = keys[i];
}
}
if (max == 0){ | return -1; | }
return max;
}
|
//Given a string s and a natural number n, you have been tasked to implement
// a function that returns an array of all words from string s that contain exactly
// n consonants, in order these words appear in the string s.
// If the string s is empty then the function should return an empty array.
// Note: you may assume the input string contains only letters and spaces.
// Examples:
// >>> select_words("Mary had a little lamb", 4)
// ["little"]
// >>> select_words("Mary had a little lamb", 3)
// ["Mary", "lamb"]
// >>> select_words("simple white space", 2)
// []
// >>> select_words("Hello world", 4)
// ["world"]
// >>> select_words("Uncle sam", 3)
// ["Uncle"]
function select_words(s, n){
| let words = s.split(' ').filter(e => e != ''); | let selected = words.filter(e => e.match(/[bcdfghjklmnpqrstvwxyz]/gi) != null);
if(n > 0){
return selected.filter(e => e.match(/[bcdfghjklmnpqrstvwxyz]/gi).length == n);
}else{
return [];
}
}
|
//Given a positive integer n, return the count of the numbers of n-digit
// positive integers that start or end with 1.
function starts_one_ends(n){
let ones = []; | for(let i = Math.pow(10, n-1); i < Math.pow(10, n); i++){ | if(i.toString().startsWith('1') || i.toString().endsWith('1')){
ones.push(i);
}
}
return ones.length;
} |
//Return true is array elements are monotonically increasing or decreasing.
// >>> monotonic([1, 2, 4, 20])
// true
// >>> monotonic([1, 20, 4, 10])
// false
// >>> monotonic([4, 1, 0, -10])
// true
function monotonic(l){
if(l.length <= 1){
return true;
}
if(l[0] < l[1]){
for(var i = 1; i < l.length - 1; i++){
if(l[i] > l[i + 1]){
return false; | } | }
}
else{
for(var i = 1; i < l.length - 1; i++){
if(l[i] < l[i + 1]){
return false;
}
}
}
return true;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Checks if given string is a palindrome
// >>> isPalindrome((""))
// (true)
// >>> isPalindrome(("aba"))
// (true)
// >>> isPalindrome(("aaaaa"))
// (true)
// >>> isPalindrome(("zbcd"))
// (false)
public static boolean isPalindrome(String text) {
boolean isPalindrome = true;
int left = 0;
int right = text.length() - 1;
while (left < right) {
if (text.charAt(left) != text.charAt(right)) { | isPalindrome = false; | break;
}
left++;
right--;
}
return isPalindrome;
}
} |
from typing import List
def exchange(lst1: List[int], lst2: List[int]) -> str:
"""In this problem, you will implement a function that takes two lists of numbers,
and determines whether it is possible to perform an exchange of elements
between them to make lst1 a list of only even numbers.
There is no limit on the number of exchanged elements between lst1 and lst2.
If it is possible to exchange elements between the lst1 and lst2 to make
all the elements of lst1 to be even, return "YES".
Otherwise, return "NO".
For example:
>>> exchange([1, 2, 3, 4], [1, 2, 3, 4])
'YES'
>>> exchange([1, 2, 3, 4], [1, 5, 3, 4])
'NO'
It is assumed that the input lists will be non-empty.
"""
| count = 0 | for x in lst1:
if x % 2 == 1:
count += 1
for y in lst2:
if y % 2 == 0:
count -= 1
if count > 0:
return "NO"
else:
return "YES" |
//Check if in given array of numbers, are any two numbers closer to each other than
// given threshold.
// >>> has_close_elements([1.0, 2.0, 3.0], 0.5)
// false
// >>> has_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)
// true
function has_close_elements(numbers, threshold){
for(var i = 0; i < numbers.length; i++){
for(var j = i + 1; j < numbers.length; j++){
if(Math.abs(numbers[i] - numbers[j]) < threshold){
return true;
} | } | }
return false;
}
|
from typing import List
def intersperse(numbers: List[int], delimeter: int) -> List[int]:
""" Insert a number 'delimeter' between every two consecutive elements of input list `numbers'
>>> intersperse([], 4)
[]
>>> intersperse([1, 2, 3], 4)
[1, 4, 2, 4, 3]
"""
result = []
if len(numbers) == 0: | return result | result.append(numbers[0])
for num in numbers[1:]:
result.append(delimeter)
result.append(num)
return result
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// pairs_sum_to_zero takes an array array list of integers as an input.
// it returns true if there are two distinct elements in the array list that
// sum to zero, and false otherwise.
// >>> pairsSumToZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)5l, (long)0l))))
// (false)
// >>> pairsSumToZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)-2l, (long)1l))))
// (false)
// >>> pairsSumToZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)7l))))
// (false)
// >>> pairsSumToZero((new ArrayList<Long>(Arrays.asList((long)2l, (long)4l, (long)-5l, (long)3l, (long)5l, (long)7l))))
// (true)
// >>> pairsSumToZero((new ArrayList<Long>(Arrays.asList((long)1l))))
// (false)
public static boolean pairsSumToZero(ArrayList<Long> l) {
for (int i = 0; i < l.size(); i++) {
for (int j = i + 1; j < l.size(); j++) {
if (l.get(i) + l.get(j) == 0) {
return true;
} | } | }
return false;
}
} |
//Given a string representing a space separated lowercase letters, return an object
// of the letter with the most repetition and containing the corresponding count.
// If several letters have the same occurrence, return all of them.
// Example:
// >>> histogram("a b c")
// {"a": 1, "b": 1, "c": 1}
// >>> histogram("a b b a")
// {"a": 2, "b": 2}
// >>> histogram("a b c a b")
// {"a": 2, "b": 2}
// >>> histogram("b b b b a")
// {"b": 4}
// >>> histogram("")
// {}
function histogram(test){
if (test.length === 0){
return {};
}
var lst = test.split(" ");
var count = {};
for (var i = 0; i < lst.length; i++){
if (count[lst[i]]){
count[lst[i]] += 1;
}
else{
count[lst[i]] = 1;
}
}
console.log(count);
var max = 0;
var result = {};
for (var key in count){
if (count[key] > max){
max = count[key];
result = {};
result[key] = max;
}
else if (count[key] === max){ | result[key] = max; | }
}
return result;
}
|
from typing import List, Tuple, Optional
def largest_smallest_integers(lst: List[int]) -> Tuple[Optional[int], Optional[int]]:
"""
Create a function that returns a tuple (a, b), where 'a' is
the largest of negative integers, and 'b' is the smallest
of positive integers in a list.
If there is no negative or positive integers, return them as None.
Examples:
>>> largest_smallest_integers([2, 4, 1, 3, 5, 7])
(None, 1)
>>> largest_smallest_integers([])
(None, None)
>>> largest_smallest_integers([0])
(None, None)
"""
smallest_positive = None
largest_negative = None
for num in lst: | if num > 0: | if smallest_positive is None or num < smallest_positive:
smallest_positive = num
elif num < 0:
if largest_negative is None or num > largest_negative:
largest_negative = num
return (largest_negative, smallest_positive)
|
//Given a positive integer n, return the product of the odd digits.
// Return 0 if all digits are even.
// For example:
// >>> digits(1)
// 1
// >>> digits(4)
// 0
// >>> digits(235)
// 15
function digits(n){
if(!n){
return "error";
}
if(n < 0){
return "error";
}
var temp = 0;
var prod = 1;
var odd = true;
for(n; n > 0; n = Math.floor(n/10)){
temp = n % 10;
if(temp % 2 != 0){
prod *= temp; | odd = false; | }
}
if(odd){
return 0;
}
return prod;
}
|
//You are given a non-empty array of positive integers. Return the greatest integer that is greater than
// zero, and has a frequency greater than or equal to the value of the integer itself.
// The frequency of an integer is the number of times it appears in the array.
// If no such a value exist, return -1.
// Examples:
// >>> search([4, 1, 2, 2, 3, 1])
// 2
// >>> search([1, 2, 2, 3, 3, 3, 4, 4, 4])
// 3
// >>> search([5, 5, 4, 4, 4])
// -1
function search(lst){
var h = {};
for (var i=0; i<lst.length; i++){
var x = lst[i];
if (x in h){
h[x] += 1;
}
else{
h[x] = 1;
}
}
var keys = Object.keys(h);
var max = 0;
for (var i=0; i<keys.length; i++){
if (keys[i] > max && h[keys[i]] >= keys[i]){
max = keys[i];
} | } | if (max == 0){
return -1;
}
return max;
}
|
//Given an array of numbers, return whether or not they are sorted
// in ascending order. If array has more than 1 duplicate of the same
// number, return false. Assume no negative numbers and only integers.
// Examples
// >>> is_sorted([5])
// true
// >>> is_sorted([1, 2, 3, 4, 5])
// true
// >>> is_sorted([1, 3, 2, 4, 5])
// false
// >>> is_sorted([1, 2, 3, 4, 5, 6])
// true
// >>> is_sorted([1, 2, 3, 4, 5, 6, 7])
// true
// >>> is_sorted([1, 3, 2, 4, 5, 6, 7])
// false
// >>> is_sorted([1, 2, 2, 3, 3, 4])
// true
// >>> is_sorted([1, 2, 2, 2, 3, 4])
// false
function is_sorted(lst){
var last = null;
var dupe_found = false;
for (var x of lst){
if (last == null){
last = x;
} else if (x < last){
return false;
} else if (x == last){
if (dupe_found){
return false;
} else {
dupe_found = true;
} | } else { | last = x;
dupe_found = false;
}
}
return true;
}
|
def car_race_collision(n: int) -> int:
"""
Imagine a road that's a perfectly straight infinitely long line.
n cars are driving left to right; simultaneously, a different set of n cars
are driving right to left. The two sets of cars start out being very far from
each other. All cars move in the same speed. Two cars are said to collide
when a car that's moving left to right hits a car that's moving right to left.
However, the cars are infinitely sturdy and strong; as a result, they continue moving
in their trajectory as if they did not collide.
This function outputs the number of such collisions.
"""
l = n * n
count = 0 | left_cars = [i for i in range(l)] | right_cars = [i for i in range(l)]
for i in range(l):
for j in range(l):
if left_cars[i] == right_cars[j]:
count += 1
return count |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// This function takes an array array list l and returns an array array list l' such that
// l' is identical to l in the indicies that are not divisible by three, while its values at the indicies that are divisible by three are equal
// to the values of the corresponding indicies of l, but sorted.
// >>> sortThird((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l))))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l)))
// >>> sortThird((new ArrayList<Long>(Arrays.asList((long)5l, (long)6l, (long)3l, (long)4l, (long)8l, (long)9l, (long)2l))))
// (new ArrayList<Long>(Arrays.asList((long)2l, (long)6l, (long)3l, (long)4l, (long)8l, (long)9l, (long)5l)))
public static ArrayList<Long> sortThird(ArrayList<Long> l) {
ArrayList<Long> l_r = new ArrayList<Long>();
ArrayList<Long> l_e = new ArrayList<Long>();
ArrayList<Long> l_s = new ArrayList<Long>();
for (int i = 0; i < l.size(); i++) {
if (i % 3 == 0) {
l_e.add(l.get(i));
} else {
l_r.add(l.get(i));
}
}
for (int i = 0; i < l_e.size(); i++) {
long min = l_e.get(i);
for (int j = i; j < l_e.size(); j++) {
if (l_e.get(j) < min) { | min = l_e.get(j); | l_e.set(j, l_e.get(i));
l_e.set(i, min);
}
}
}
int j = 0;
for (int i = 0; i < l.size(); i++) {
if (i % 3 == 0) {
l_s.add(l_e.get(j));
j++;
} else {
l_s.add(l_r.get(i - j));
}
}
return l_s;
}
} |
//Write a function that takes an integer a and returns true
// if this ingeger is a cube of some integer number.
// Note: you may assume the input is always valid.
// Examples:
// >>> iscube(1)
// true
// >>> iscube(2)
// false
// >>> iscube(-1)
// true
// >>> iscube(64)
// true
// >>> iscube(0)
// true
// >>> iscube(180)
// false
function iscube(a){
if(Math.cbrt(a)%1==0){
return true; | }else{ | return false;
}
} |
from typing import List
def numerical_letter_grade(grades: List[float]) -> List[str]:
"""It is the last week of the semester and the teacher has to give the grades
to students. The teacher has been making her own algorithm for grading.
The only problem is, she has lost the code she used for grading.
She has given you a list of GPAs for some students and you have to write
a function that can output a list of letter grades using the following table:
GPA | Letter grade
4.0 A+
> 3.7 A
> 3.3 A-
> 3.0 B+
> 2.7 B
> 2.3 B-
> 2.0 C+
> 1.7 C
> 1.3 C-
> 1.0 D+
> 0.7 D
> 0.0 D-
0.0 E
Example:
>>> grade_equation([4.0, 3, 1.7, 2, 3.5])
['A+', 'B', 'C-', 'C', 'A-']
"""
final = []
for x in grades:
if x == 4.0:
final.append('A+')
elif x > 3.7:
final.append('A')
elif x > 3.3:
final.append('A-')
elif x > 3.0:
final.append('B+')
elif x > 2.7:
final.append('B')
elif x > 2.3:
final.append('B-')
elif x > 2.0:
final.append('C+')
elif x > 1.7:
final.append('C')
elif x > 1.3:
final.append('C-')
elif x > 1.0:
final.append('D+')
elif x > 0.7:
final.append('D')
elif x > 0.0:
final.append('D-')
else: | final.append('E') | return final
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// You're a hungry rabbit, and you already have eaten a certain number of carrots,
// but now you need to eat more carrots to complete the day's meals.
// you should return an array array list of [ total number of eaten carrots after your meals,
// the number of carrots left after your meals ]
// if there are not enough remaining carrots, you will eat all remaining carrots, but will still be hungry.
// Example:
// >>> eat((5l), (6l), (10l))
// (new ArrayList<Long>(Arrays.asList((long)11l, (long)4l)))
// >>> eat((4l), (8l), (9l))
// (new ArrayList<Long>(Arrays.asList((long)12l, (long)1l)))
// >>> eat((1l), (10l), (10l))
// (new ArrayList<Long>(Arrays.asList((long)11l, (long)0l)))
// >>> eat((2l), (11l), (5l))
// (new ArrayList<Long>(Arrays.asList((long)7l, (long)0l)))
// Variables:
// @number : integer
// the number of carrots that you have eaten.
// @need : integer
// the number of carrots that you need to eat.
// @remaining : integer
// the number of remaining carrots thet exist in stock
// Constrain:
// * 0 <= number <= 1000
// * 0 <= need <= 1000
// * 0 <= remaining <= 1000
// Have fun :)
public static ArrayList<Long> eat(long number, long need, long remaining) {
long total = number + Math.min(need, remaining);
long left = Math.max(0, remaining - need);
return new ArrayList<Long>(Arrays.asList(total, left));
} | } | |
//There are eight planets in our solar system: the closerst to the Sun
// is Mercury, the next one is Venus, then Earth, Mars, Jupiter, Saturn,
// Uranus, Neptune.
// Write a function that takes two planet names as strings planet1 and planet2.
// The function should return an array containing all planets whose orbits are
// located between the orbit of planet1 and the orbit of planet2, sorted by
// the proximity to the sun.
// The function should return an empty array if planet1 or planet2
// are not correct planet names.
// Examples
// >>> bf("Jupiter", "Neptune")
// ["Saturn", "Uranus"]
// >>> bf("Earth", "Mercury")
// "Venus"
// >>> bf("Mercury", "Uranus")
// ["Venus", "Earth", "Mars", "Jupiter", "Saturn"]
function bf(planet1, planet2){
var planets = {
'Mercury':1,
'Venus':2,
'Earth':3,
'Mars':4,
'Jupiter':5,
'Saturn':6,
'Uranus':7,
'Neptune':8
}
if(!planets[planet1] || !planets[planet2]){
return [];
}
var keys = Object.keys(planets);
var position1 = keys.indexOf(planet1);
var position2 = keys.indexOf(planet2); | if(position1 > position2){ | var temp = position1;
position1 = position2;
position2 = temp;
}
var res = [];
for(var i = position1 + 1; i < position2; i++){
res.push(keys[i]);
}
return res;
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Circular shift the digits of the integer x, shift the digits right by shift
// and return the result as a string.
// If shift > number of digits, return digits reversed.
// >>> circularShift((12l), (1l))
// ("21")
// >>> circularShift((12l), (2l))
// ("12")
public static String circularShift(long x, long shift) {
String digits = String.valueOf(x);
if (shift > digits.length()) {
return new StringBuilder(digits).reverse().toString();
} | return (digits.substring(digits.length() - (int)shift, digits.length()) + digits.substring(0, digits.length() - (int)shift)); | }
} |
def int_to_mini_roman(number: int) -> str:
"""
Given a positive integer, obtain its roman numeral equivalent as a string,
and return it in lowercase.
Restrictions: 1 <= num <= 1000
Examples:
>>> int_to_mini_roman(19)
'xix'
>>> int_to_mini_roman(152)
'clii'
>>> int_to_mini_roman(426)
'cdxxvi'
"""
roman_numerals = {
1000: 'm',
900: 'cm',
500: 'd',
400: 'cd',
100: 'c',
90: 'xc',
50: 'l',
40: 'xl', | 10: 'x', | 9: 'ix',
5: 'v',
4: 'iv',
1: 'i'
}
roman_numeral = ''
for key in sorted(roman_numerals.keys(), reverse=True):
roman_numeral += roman_numerals[key] * (number // key)
number = number % key
return roman_numeral
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// You are given an array array list of integers.
// You need to find the largest prime value and return the sum of its digits.
// Examples:
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)3l, (long)2l, (long)1l, (long)3l, (long)5l, (long)7l, (long)4l, (long)5l, (long)5l, (long)5l, (long)2l, (long)181l, (long)32l, (long)4l, (long)32l, (long)3l, (long)2l, (long)32l, (long)324l, (long)4l, (long)3l))))
// (10l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)1l, (long)0l, (long)1l, (long)8l, (long)2l, (long)4597l, (long)2l, (long)1l, (long)3l, (long)40l, (long)1l, (long)2l, (long)1l, (long)2l, (long)4l, (long)2l, (long)5l, (long)1l))))
// (25l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)1l, (long)32l, (long)5107l, (long)34l, (long)83278l, (long)109l, (long)163l, (long)23l, (long)2323l, (long)32l, (long)30l, (long)1l, (long)9l, (long)3l))))
// (13l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)724l, (long)32l, (long)71l, (long)99l, (long)32l, (long)6l, (long)0l, (long)5l, (long)91l, (long)83l, (long)0l, (long)5l, (long)6l))))
// (11l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)81l, (long)12l, (long)3l, (long)1l, (long)21l))))
// (3l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)8l, (long)1l, (long)2l, (long)1l, (long)7l))))
// (7l)
public static long skjkasdkd(ArrayList<Long> lst) {
long largestPrime = 0l;
for (Long item : lst) {
long number = item;
int counter = 0;
if (item == 1) {
continue;
}
if (item == 2) {
counter = 1;
}
else {
int l = 1;
while (l <= (int) number) {
if (number % l == 0) {
counter++;
}
l++;
}
}
if (counter == 2) {
if (number > largestPrime) {
largestPrime = number;
}
} | } | if (largestPrime == 0) {
return 0l;
}
else {
long output = 0l;
while (largestPrime > 0) {
output += largestPrime % 10;
largestPrime /= 10;
}
return output;
}
}
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given a string representing a space separated lowercase letters, return a hash map
// of the letter with the most repetition and containing the corresponding count.
// If several letters have the same occurrence, return all of them.
// Example:
// >>> histogram(("a b c"))
// (new HashMap<String,Long>(Map.of("a", 1l, "b", 1l, "c", 1l)))
// >>> histogram(("a b b a"))
// (new HashMap<String,Long>(Map.of("a", 2l, "b", 2l)))
// >>> histogram(("a b c a b"))
// (new HashMap<String,Long>(Map.of("a", 2l, "b", 2l)))
// >>> histogram(("b b b b a"))
// (new HashMap<String,Long>(Map.of("b", 4l)))
// >>> histogram((""))
// (new HashMap<String,Long>())
public static HashMap<String,Long> histogram(String test) {
HashMap<String,Long> s = new HashMap<String,Long>();
if (test.length() > 0) { | HashMap<String,Long> hm = new HashMap<String,Long>(); | String[] ss = test.split(" ");
for (String sss : ss) {
if (hm.containsKey(sss)) {
hm.put(sss, hm.get(sss) + 1);
} else {
hm.put(sss, 1l);
}
}
long max = 0;
for (String key : hm.keySet()) {
if (hm.get(key) > max) {
max = hm.get(key);
}
}
for (String key : hm.keySet()) {
if (hm.get(key) == max) {
s.put(key, hm.get(key));
}
}
}
return s;
}
} |
//Create a function that takes a value (string) representing a number
// and returns the closest integer to it. If the number is equidistant
// from two integers, round it away from zero.
// Examples
// >>> closest_integer("10")
// 10
// >>> closest_integer("15.3")
// 15
// Note:
// Rounding away from zero means that if the given number is equidistant
// from two integers, the one you should return is the one that is the
// farthest from zero. For example closest_integer("14.5") should
// return 15 and closest_integer("-14.5") should return -15.
function closest_integer(value){
value = parseFloat(value);
if(value > 0){
if(value - Math.floor(value) < 0.5){
return Math.floor(value);
}else{
return Math.ceil(value);
}
}else{
if(value - Math.floor(value) > -0.5){
return Math.floor(value);
}else{
return Math.ceil(value);
} | } | }
|
def is_multiply_prime(a: int) -> bool:
"""Write a function that returns true if the given number is the multiplication of 3 prime numbers
and false otherwise.
Knowing that (a) is less then 100.
Example:
>>> is_multiply_prime(30)
True
30 = 2 * 3 * 5
"""
n = a
d = 2 | prime_factors = [] | while d * d <= n:
if n % d == 0:
prime_factors.append(d)
n //= d
else:
d += 1
if n > 1:
prime_factors.append(n)
if len(prime_factors) == 3:
return True
else:
return False
|
from typing import Union
def compare_one(a: Union[int, float, str], b: Union[int, float, str]) -> Union[int, float, str, None]:
"""
Create a function that takes integers, floats, or strings representing
real numbers, and returns the larger variable in its given variable type.
Return None if the values are equal.
Note: If a real number is represented as a string, the floating point might be . or ,
>>> compare_one(1, 2.5)
2.5
>>> compare_one(1, '2,3')
'2,3'
>>> compare_one('5,1', '6')
'6'
>>> compare_one('1', 1)
None
"""
if isinstance(a, int) and isinstance(b, int):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, float) and isinstance(b, float):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, str) and isinstance(b, str):
a_tmp = a.replace(',', '.')
b_tmp = b.replace(',', '.') | if float(a_tmp) > float(b_tmp): | return a
elif float(a_tmp) < float(b_tmp):
return b
else:
return None
elif isinstance(a, int) and isinstance(b, float):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, float) and isinstance(b, int):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, str) and isinstance(b, int):
a_tmp = a.replace(',', '.')
if float(a_tmp) > b:
return a
elif float(a_tmp) < b:
return b
else:
return None
elif isinstance(a, int) and isinstance(b, str):
b_tmp = b.replace(',', '.')
if a > float(b_tmp):
return a
elif a < float(b_tmp):
return b
else:
return None
elif isinstance(a, str) and isinstance(b, float):
a_tmp = a.replace(',', '.')
if float(a_tmp) > b:
return a
elif float(a_tmp) < b:
return b |
from typing import Union, List
def split_words(txt: str) -> Union[List[str], int]:
"""
Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
Examples
>>> split_words('Hello world!')
['Hello', 'world!']
>>> split_words('Hello,world!')
['Hello', 'world!']
>>> split_words('abcdef')
3
"""
alphabet = [chr(letter) for letter in range(ord('a'), ord('z') + 1)]
if ' ' in txt:
return txt.split(' ')
elif ',' in txt:
return txt.split(',')
else: | return len([letter for letter in txt if letter in alphabet and alphabet.index(letter) % 2 == 1]) | |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Input to this function is a string containing multiple groups of nested parentheses. Your goal is to
// separate those group into separate strings and return the array list of those.
// Separate groups are balanced (each open brace is properly closed) and not nested within each other
// Ignore any spaces in the input string.
// >>> separateParenGroups(("( ) (( )) (( )( ))"))
// (new ArrayList<String>(Arrays.asList((String)"()", (String)"(())", (String)"(()())")))
public static ArrayList<String> separateParenGroups(String paren_string) {
ArrayList<String> output = new ArrayList<String>();
StringBuffer temp = new StringBuffer();
int open_paren = 0;
for (int i = 0; i < paren_string.length(); i++) {
char c = paren_string.charAt(i);
if (c == '(') {
if (open_paren == 0) {
temp = new StringBuffer();
}
open_paren++;
temp.append(c);
} else if (c == ')') {
open_paren--;
temp.append(c);
if (open_paren == 0) {
output.add(temp.toString());
} | } | }
return output;
}
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// xs represent coefficients of a polynomial.
// xs[0] + xs[1] * x + xs[2] * x^2 + ....
// Return derivative of this polynomial in the same form.
// >>> derivative((new ArrayList<Long>(Arrays.asList((long)3l, (long)1l, (long)2l, (long)4l, (long)5l))))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)4l, (long)12l, (long)20l)))
// >>> derivative((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l))))
// (new ArrayList<Long>(Arrays.asList((long)2l, (long)6l)))
public static ArrayList<Long> derivative(ArrayList<Long> xs) {
ArrayList<Long> ans = new ArrayList<Long>();
for (int i = 1; i < xs.size(); i++) {
ans.add(xs.get(i) * i); | } | return ans;
}
} |
def fizz_buzz(n: int) -> int:
"""Return the number of times the digit 7 appears in integers less than n which are divisible by 11 or 13.
>>> fizz_buzz(50)
0
>>> fizz_buzz(78)
2
>>> fizz_buzz(79)
3
"""
x = 0
for i in range(n): | if i % 11 == 0 or i % 13 == 0: | x += str(i).count('7')
return x
|
def vowels_count(s: str) -> int:
"""Write a function vowels_count which takes a string representing
a word as input and returns the number of vowels in the string.
Vowels in this case are 'a', 'e', 'i', 'o', 'u'. Here, 'y' is also a
vowel, but only when it is at the end of the given word.
Example:
>>> vowels_count('abcde')
2
>>> vowels_count('ACEDY')
3
"""
s = s.lower()
i = 0
count = 0
while i < len(s):
if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i' or s[i] == 'o' or s[i] == 'u'):
count += 1 | if (s[i] == 'y' and i == len(s) - 1): | count += 1
i += 1
return count
|
def iscube(a: int) -> bool:
"""
Write a function that takes an integer a and returns True
if this ingeger is a cube of some integer number.
Note: you may assume the input is always valid.
Examples:
>>> iscube(1)
True
>>> iscube(2)
False
>>> iscube(-1)
True
>>> iscube(64)
True
>>> iscube(0)
True
>>> iscube(180)
False
"""
import math | a = abs(a) | if a == 0:
return True
for i in range(1,int(math.sqrt(a))+1):
if i**3 == a:
return True
return False
pass
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// You are given a positive integer n. You have to create an integer array array list a of length n.
// For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.
// Return the number of triples (a[i], a[j], a[k]) of a where i < j < k,
// and a[i] + a[j] + a[k] is a multiple of 3.
// Example :
// >>> getMaxTriples((5l))
// (1l)
// Explanation:
// a = [1, 3, 7, 13, 21]
// The only valid triple is (1, 7, 13).
public static long getMaxTriples(long n) {
int count = 0;
ArrayList<Integer> arrayList = new ArrayList<Integer>();
for (int i = 1; i <= n; i++) {
arrayList.add(i * i - i + 1);
}
for (int i = 0; i < arrayList.size(); i++) {
for (int j = i + 1; j < arrayList.size(); j++) {
for (int k = j + 1; k < arrayList.size(); k++) { | if ((arrayList.get(i) + arrayList.get(j) + arrayList.get(k)) % 3 == 0) { | count++;
}
}
}
}
return count;
}
} |
//From a given array of integers, generate an array of rolling maximum element found until given moment
// in the sequence.
// >>> rolling_max([1, 2, 3, 2, 3, 4, 2])
// [1, 2, 3, 3, 3, 4, 4]
function rolling_max(numbers){
let max = [];
let tempMax = 0;
for(let i = 0; i < numbers.length; i++){ | if(tempMax < numbers[i]){ | tempMax = numbers[i];
}
max.push(tempMax);
}
return max;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given an array array list arr of integers and a positive integer k, return a sorted array list
// of length k with the maximum k numbers in arr.
// Example 1:
// >>> maximum((new ArrayList<Long>(Arrays.asList((long)-3l, (long)-4l, (long)5l))), (3l))
// (new ArrayList<Long>(Arrays.asList((long)-4l, (long)-3l, (long)5l)))
// Example 2:
// >>> maximum((new ArrayList<Long>(Arrays.asList((long)4l, (long)-4l, (long)4l))), (2l))
// (new ArrayList<Long>(Arrays.asList((long)4l, (long)4l)))
// Example 3:
// >>> maximum((new ArrayList<Long>(Arrays.asList((long)-3l, (long)2l, (long)1l, (long)2l, (long)-1l, (long)-2l, (long)1l))), (1l))
// (new ArrayList<Long>(Arrays.asList((long)2l)))
// Note:
// 1. The length of the array array list will be in the range of [1, 1000].
// 2. The elements in the array array list will be in the range of [-1000, 1000].
// 3. 0 <= k <= len(arr)
public static ArrayList<Long> maximum(ArrayList<Long> arr, long k) {
Collections.sort(arr);
return new ArrayList<Long>(arr.subList((int)(arr.size() - k), (int)(arr.size()))); | } | } |
//Write a function that accepts an array of strings.
// The array contains different words. Return the word with maximum number
// of unique characters. If multiple strings have maximum number of unique
// characters, return the one which comes first in lexicographical order.
// >>> find_max(["name", "of", "string"])
// "string"
// >>> find_max(["name", "enam", "game"])
// "enam"
// >>> find_max(["aaaaaaa", "bb", "cc"])
// "aaaaaaa"
function find_max(words){
var max_unique = {};
var max_unique_word = "";
for(var i = 0; i < words.length; i++){
var unique_word = {};
var word = words[i];
for(var j = 0; j < word.length; j++){
var char = word[j];
if(unique_word[char] === undefined){ | unique_word[char] = 1; | }
}
if(max_unique_word === ""){
max_unique_word = word;
max_unique = unique_word;
}
else if(Object.keys(unique_word).length > Object.keys(max_unique).length){
max_unique_word = word;
max_unique = unique_word;
}
else if(Object.keys(unique_word).length === Object.keys(max_unique).length){
if(word < max_unique_word){
max_unique_word = word;
max_unique = unique_word;
}
}
}
return max_unique_word;
} |
from typing import Union
def compare_one(a: Union[int, float, str], b: Union[int, float, str]) -> Union[int, float, str, None]:
"""
Create a function that takes integers, floats, or strings representing
real numbers, and returns the larger variable in its given variable type.
Return None if the values are equal.
Note: If a real number is represented as a string, the floating point might be . or ,
>>> compare_one(1, 2.5)
2.5
>>> compare_one(1, '2,3')
'2,3'
>>> compare_one('5,1', '6')
'6'
>>> compare_one('1', 1)
None
"""
if isinstance(a, int) and isinstance(b, int):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, float) and isinstance(b, float): | if a > b: | return a
elif a < b:
return b
else:
return None
elif isinstance(a, str) and isinstance(b, str):
a_tmp = a.replace(',', '.')
b_tmp = b.replace(',', '.')
if float(a_tmp) > float(b_tmp):
return a
elif float(a_tmp) < float(b_tmp):
return b
else:
return None
elif isinstance(a, int) and isinstance(b, float):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, float) and isinstance(b, int):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, str) and isinstance(b, int):
a_tmp = a.replace(',', '.')
if float(a_tmp) > b:
return a
elif float(a_tmp) < b:
return b
else:
return None
elif isinstance(a, int) and isinstance(b, str):
b_tmp = b.replace(',', '.')
if a > float(b_tmp):
return a
elif a < float(b_tmp):
return b
else:
return None
elif isinstance(a, str) and isinstance(b, float):
a_tmp = a.replace(',', '.')
if float(a_tmp) > b:
return a
elif float(a_tmp) < b:
return b |
//Given a grid with N rows and N columns (N >= 2) and a positive integer k,
// each cell of the grid contains a value. Every integer in the range [1, N * N]
// inclusive appears exactly once on the cells of the grid.
// You have to find the minimum path of length k in the grid. You can start
// from any cell, and in each step you can move to any of the neighbor cells,
// in other words, you can go to cells which share an edge with you current
// cell.
// Please note that a path of length k means visiting exactly k cells (not
// necessarily distinct).
// You CANNOT go off the grid.
// A path A (of length k) is considered less than a path B (of length k) if
// after making the ordered arrays of the values on the cells that A and B go
// through (let's call them lst_A and lst_B), lst_A is lexicographically less
// than lst_B, in other words, there exist an integer index i (1 <= i <= k)
// such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
// lst_A[j] = lst_B[j].
// It is guaranteed that the answer is unique.
// Return an ordered array of the values on the cells that the minimum path go through.
// Examples:
// >>> minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3)
// [1, 2, 1]
// >>> minPath([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1)
// [1]
function minPath(grid, k){
let n = grid.length;
let path = [];
let startCell = grid[0][0];
for(let i = 0; i < n; i++){
for(let j = 0; j < n; j++){
let curr = grid[i][j];
if(curr < startCell){
startCell = curr;
} | } | }
let currCell = startCell;
for(let l = 0; l < k; l++){
path.push(currCell);
let neighbours = [];
let x = null;
let y = null;
for(let i = 0; i < n; i++){
for(let j = 0; j < n; j++){
if(grid[i][j] === currCell){
x = i;
y = j;
}
}
}
if(x > 0){
neighbours.push(grid[x - 1][y]);
}
if(x < n - 1){
neighbours.push(grid[x + 1][y]);
}
if(y > 0){
neighbours.push(grid[x][y - 1]);
}
if(y < n - 1){
neighbours.push(grid[x][y + 1]);
}
let nextCell = neighbours[0];
for(let i = 0; i < neighbours.length; i++){
let curr = neighbours[i];
if(curr < nextCell){
nextCell = curr;
}
}
currCell = nextCell;
}
return path;
} |
//Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers
// Example
// >>> is_equal_to_sum_even(4)
// false
// >>> is_equal_to_sum_even(6)
// false
// >>> is_equal_to_sum_even(8)
// true
function is_equal_to_sum_even(n){
let m =0;
for(let i=2;i<n;i+=2){
for(let j=2;j<n;j+=2){
for(let k=2;k<n;k+=2){
for(let l=2;l<n;l+=2){
if((i+j+k+l)==n){ | m=1; | break;
}
}
}
}
}
if(m==1){
return true;
}
else{
return false;
}
} |
//Your task is to implement a function that will simplify the expression
// x * n. The function returns true if x * n evaluates to a whole number and false
// otherwise. Both x and n, are string representation of a fraction, and have the following format,
// <numerator>/<denominator> where both numerator and denominator are positive whole numbers.
// You can assume that x, and n are valid fractions, and do not have zero as denominator.
// >>> simplify("1/5", "5/1")
// true
// >>> simplify("1/6", "2/1")
// false
// >>> simplify("7/10", "10/2")
// false
function simplify(x, n){
| let xNumerator = parseInt(x.split('/')[0]); | let xDenominator = parseInt(x.split('/')[1]);
let nNumerator = parseInt(n.split('/')[0]);
let nDenominator = parseInt(n.split('/')[1]);
let xTimesN = xNumerator * nNumerator / (xDenominator * nDenominator);
return Number.isInteger(xTimesN);
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// This function takes an array array list l and returns an array array list l' such that
// l' is identical to l in the indicies that are not divisible by three, while its values at the indicies that are divisible by three are equal
// to the values of the corresponding indicies of l, but sorted.
// >>> sortThird((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l))))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l)))
// >>> sortThird((new ArrayList<Long>(Arrays.asList((long)5l, (long)6l, (long)3l, (long)4l, (long)8l, (long)9l, (long)2l))))
// (new ArrayList<Long>(Arrays.asList((long)2l, (long)6l, (long)3l, (long)4l, (long)8l, (long)9l, (long)5l)))
public static ArrayList<Long> sortThird(ArrayList<Long> l) {
ArrayList<Long> l_r = new ArrayList<Long>();
ArrayList<Long> l_e = new ArrayList<Long>();
ArrayList<Long> l_s = new ArrayList<Long>();
for (int i = 0; i < l.size(); i++) {
if (i % 3 == 0) {
l_e.add(l.get(i));
} else {
l_r.add(l.get(i));
} | } | for (int i = 0; i < l_e.size(); i++) {
long min = l_e.get(i);
for (int j = i; j < l_e.size(); j++) {
if (l_e.get(j) < min) {
min = l_e.get(j);
l_e.set(j, l_e.get(i));
l_e.set(i, min);
}
}
}
int j = 0;
for (int i = 0; i < l.size(); i++) {
if (i % 3 == 0) {
l_s.add(l_e.get(j));
j++;
} else {
l_s.add(l_r.get(i - j));
}
}
return l_s;
}
} |
//Given a grid with N rows and N columns (N >= 2) and a positive integer k,
// each cell of the grid contains a value. Every integer in the range [1, N * N]
// inclusive appears exactly once on the cells of the grid.
// You have to find the minimum path of length k in the grid. You can start
// from any cell, and in each step you can move to any of the neighbor cells,
// in other words, you can go to cells which share an edge with you current
// cell.
// Please note that a path of length k means visiting exactly k cells (not
// necessarily distinct).
// You CANNOT go off the grid.
// A path A (of length k) is considered less than a path B (of length k) if
// after making the ordered arrays of the values on the cells that A and B go
// through (let's call them lst_A and lst_B), lst_A is lexicographically less
// than lst_B, in other words, there exist an integer index i (1 <= i <= k)
// such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we have
// lst_A[j] = lst_B[j].
// It is guaranteed that the answer is unique.
// Return an ordered array of the values on the cells that the minimum path go through.
// Examples:
// >>> minPath([[1, 2, 3], [4, 5, 6], [7, 8, 9]], 3)
// [1, 2, 1]
// >>> minPath([[5, 9, 3], [4, 1, 6], [7, 8, 2]], 1)
// [1]
function minPath(grid, k){
let n = grid.length;
let path = [];
let startCell = grid[0][0];
for(let i = 0; i < n; i++){
for(let j = 0; j < n; j++){
let curr = grid[i][j];
if(curr < startCell){
startCell = curr;
}
}
}
let currCell = startCell;
for(let l = 0; l < k; l++){
path.push(currCell);
let neighbours = [];
let x = null;
let y = null;
for(let i = 0; i < n; i++){
for(let j = 0; j < n; j++){
if(grid[i][j] === currCell){
x = i;
y = j;
}
}
}
if(x > 0){
neighbours.push(grid[x - 1][y]);
}
if(x < n - 1){
neighbours.push(grid[x + 1][y]);
}
if(y > 0){
neighbours.push(grid[x][y - 1]);
}
if(y < n - 1){
neighbours.push(grid[x][y + 1]);
} | let nextCell = neighbours[0]; | for(let i = 0; i < neighbours.length; i++){
let curr = neighbours[i];
if(curr < nextCell){
nextCell = curr;
}
}
currCell = nextCell;
}
return path;
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Return true if a given number is prime, and false otherwise.
// >>> isPrime((6l))
// (false)
// >>> isPrime((101l))
// (true)
// >>> isPrime((11l))
// (true)
// >>> isPrime((13441l))
// (true)
// >>> isPrime((61l))
// (true)
// >>> isPrime((4l))
// (false)
// >>> isPrime((1l))
// (false)
public static boolean isPrime(long n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
long i = 5;
while (i * i <= n) {
if (n % i == 0 || n % (i + 2) == 0) return false;
i += 6;
}
return true;
} | } | |
//prime_fib returns n-th number that is a Fibonacci number and it's also prime.
// >>> prime_fib(1)
// 2
// >>> prime_fib(2)
// 3
// >>> prime_fib(3)
// 5
// >>> prime_fib(4)
// 13
// >>> prime_fib(5)
// 89
function prime_fib(n){
function isPrime(number){
if(number <= 3){
return number > 1;
}
else if(number % 2 == 0 || number % 3 == 0){
return false;
}
else{
for(var i = 5; i * i <= number; i += 6){
if(number % i == 0 || number % (i + 2) == 0){
return false;
}
}
return true;
}
}
function fib(num){
var a = 1, b = 0, temp;
while (num >= 0){
temp = a;
a = a + b; | b = temp; | num--;
}
return b;
}
var res = 0;
var fibNum = 0;
while(res < n){
if(isPrime(fib(fibNum))){
res++;
}
fibNum++;
}
return fib(fibNum - 1);
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// triples_sum_to_zero takes an array array list of integers as an input.
// it returns true if there are three distinct elements in the array list that
// sum to zero, and false otherwise.
// >>> triplesSumToZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)5l, (long)0l))))
// (false)
// >>> triplesSumToZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)-2l, (long)1l))))
// (true)
// >>> triplesSumToZero((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)7l))))
// (false)
// >>> triplesSumToZero((new ArrayList<Long>(Arrays.asList((long)2l, (long)4l, (long)-5l, (long)3l, (long)9l, (long)7l))))
// (true)
// >>> triplesSumToZero((new ArrayList<Long>(Arrays.asList((long)1l))))
// (false)
public static boolean triplesSumToZero(ArrayList<Long> l) {
if (l.size() < 3) {
return false;
}
ArrayList<Long> sortedList = new ArrayList<Long>(l);
Collections.sort(sortedList);
for (int i = 0; i < sortedList.size() - 2; i++) {
int j = i + 1;
int k = sortedList.size() - 1;
while (j < k) {
long currSum = sortedList.get(i) + sortedList.get(j) + sortedList.get(k);
if (currSum == 0) {
return true;
} else if (currSum < 0) { | j++; | } else {
k--;
}
}
}
return false;
}
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Write a function that accepts an array array list of strings as a parameter,
// deletes the strings that have odd lengths from it,
// and returns the resulted array list with a sorted order,
// The array list is always an array array list of strings and never an array array list of numbers,
// and it may contain duplicates.
// The order of the array list should be ascending by length of each word, and you
// should return the array list sorted by that rule.
// If two words have the same length, sort the array list alphabetically.
// The function should return an array array list of strings in sorted order.
// You may assume that all words will have the same length.
// For example:
// >>> listSort((new ArrayList<String>(Arrays.asList((String)"aa", (String)"a", (String)"aaa"))))
// (new ArrayList<String>(Arrays.asList((String)"aa")))
// >>> listSort((new ArrayList<String>(Arrays.asList((String)"ab", (String)"a", (String)"aaa", (String)"cd"))))
// (new ArrayList<String>(Arrays.asList((String)"ab", (String)"cd")))
public static ArrayList<String> sortedListSum(ArrayList<String> lst) {
ArrayList<String> result = new ArrayList<String>();
for (String s : lst) {
if (s.length() % 2 == 0) {
result.add(s);
}
}
Collections.sort(result, new Comparator<String>() {
@Override
public int compare(String s1, String s2) {
if (s1.length() == s2.length()) {
return s1.compareTo(s2);
}
return s1.length() - s2.length();
} | }); | return result;
}
} |
from typing import List
def make_a_pile(n: int) -> List[int]:
"""
Given a positive integer n, you have to make a pile of n levels of stones.
The first level has n stones.
The number of stones in the next level is:
- the next odd number if n is odd.
- the next even number if n is even.
Return the number of stones in each level in a list, where element at index
i represents the number of stones in the level (i+1).
Examples:
>>> make_a_pile(3)
[3, 5, 7]
"""
pile = [n]
for i in range(1, n): | pile.append(pile[i-1] + 2) | return pile
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// It is the last week of the semester and the teacher has to give the grades
// to students. The teacher has been making her own algorithm for grading.
// The only problem is, she has lost the code she used for grading.
// She has given you an array array list of GPAs for some students and you have to write
// a function that can output an array array list of letter grades using the following table:
// GPA | Letter grade
// 4.0 A+
// > 3.7 A
// > 3.3 A-
// > 3.0 B+
// > 2.7 B
// > 2.3 B-
// > 2.0 C+
// > 1.7 C
// > 1.3 C-
// > 1.0 D+
// > 0.7 D
// > 0.0 D-
// 0.0 E
// Example:
// >>> gradeEquation((new ArrayList<Float>(Arrays.asList((float)4.0f, (float)3l, (float)1.7f, (float)2l, (float)3.5f))))
// (new ArrayList<String>(Arrays.asList((String)"A+", (String)"B", (String)"C-", (String)"C", (String)"A-")))
public static ArrayList<String> numericalLetterGrade(ArrayList<Float> grades) {
ArrayList<String> letterGrades = new ArrayList<String>();
for (int i = 0; i < grades.size(); i++) {
float currentGrade = grades.get(i);
if (currentGrade == 4.0f) {
letterGrades.add("A+");
} else if (currentGrade > 3.7f) {
letterGrades.add("A");
} else if (currentGrade > 3.3f) {
letterGrades.add("A-"); | } else if (currentGrade > 3.0f) { | letterGrades.add("B+");
} else if (currentGrade > 2.7f) {
letterGrades.add("B");
} else if (currentGrade > 2.3f) {
letterGrades.add("B-");
} else if (currentGrade > 2.0f) {
letterGrades.add("C+");
} else if (currentGrade > 1.7f) {
letterGrades.add("C");
} else if (currentGrade > 1.3f) {
letterGrades.add("C-");
} else if (currentGrade > 1.0f) {
letterGrades.add("D+");
} else if (currentGrade > 0.7f) {
letterGrades.add("D");
} else if (currentGrade > 0.0f) {
letterGrades.add("D-");
} else {
letterGrades.add("E");
}
}
return letterGrades;
}
} |
def is_equal_to_sum_even(n: int) -> bool:
"""Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers
Example
>>> is_equal_to_sum_even(4)
False
>>> is_equal_to_sum_even(6)
False
>>> is_equal_to_sum_even(8)
True
"""
| if n < 8: | return False
return n % 2 == 0
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given an array array list of non-negative integers, return a cojava of the given array array list after sorting,
// you will sort the given array array list in ascending order if the sum( first index value, last index value) is odd,
// or sort it in descending order if the sum( first index value, last index value) is even.
// Note:
// * don't change the given array array list.
// Examples:
// >>> sortArray((new ArrayList<Long>(Arrays.asList())))
// (new ArrayList<Long>(Arrays.asList()))
// >>> sortArray((new ArrayList<Long>(Arrays.asList((long)5l))))
// (new ArrayList<Long>(Arrays.asList((long)5l)))
// >>> sortArray((new ArrayList<Long>(Arrays.asList((long)2l, (long)4l, (long)3l, (long)0l, (long)1l, (long)5l))))
// (new ArrayList<Long>(Arrays.asList((long)0l, (long)1l, (long)2l, (long)3l, (long)4l, (long)5l)))
// >>> sortArray((new ArrayList<Long>(Arrays.asList((long)2l, (long)4l, (long)3l, (long)0l, (long)1l, (long)5l, (long)6l))))
// (new ArrayList<Long>(Arrays.asList((long)6l, (long)5l, (long)4l, (long)3l, (long)2l, (long)1l, (long)0l)))
public static ArrayList<Long> sortArray(ArrayList<Long> array) {
if(array.size()==0) {
return array;
}
if(array.size()==1) {
return array;
}
if(array.size()>1) {
if((array.get(0)+array.get(array.size()-1))%2==0) {
Collections.sort(array, Collections.reverseOrder());
}
else {
Collections.sort(array);
} | } | return array;
}
} |
def prime_length(string: str) -> bool:
"""Write a function that takes a string and returns True if the string
length is a prime number or False otherwise
Examples
>>> prime_length('Hello')
True
>>> prime_length('abcdcba')
True
>>> prime_length('kittens')
True
>>> prime_length('orange')
False
"""
num = len(string) | if num > 1: | for i in range(2, num):
if (num % i) == 0:
return False
else:
return True
else:
return False
pass
|
//Given a positive integer n, return an array that has the number of even and odd
// integer palindromes that fall within the range(1, n), inclusive.
// Example 1:
// >>> even_odd_palindrome(3)
// [1, 2]
// Explanation:
// Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.
// Example 2:
// >>> even_odd_palindrome(12)
// [4, 6]
// Explanation:
// Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.
// Note:
// 1. 1 <= n <= 10^3
// 2. returned array has the number of even and odd integer palindromes respectively.
function even_odd_palindrome(n){
var pal = [];
var isPalindrome = function(x){
var y = x.toString().split('').reverse().join('');
return y == x;
} | for(var i = 1; i <= n; i++){ | if(isPalindrome(i)){
pal.push(i);
}
}
var odd = 0;
var even = 0;
for(var i = 0; i < pal.length; i++){
if(pal[i] % 2 == 0){
even++;
}
else{
odd++;
}
}
return [even, odd];
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// It is the last week of the semester and the teacher has to give the grades
// to students. The teacher has been making her own algorithm for grading.
// The only problem is, she has lost the code she used for grading.
// She has given you an array array list of GPAs for some students and you have to write
// a function that can output an array array list of letter grades using the following table:
// GPA | Letter grade
// 4.0 A+
// > 3.7 A
// > 3.3 A-
// > 3.0 B+
// > 2.7 B
// > 2.3 B-
// > 2.0 C+
// > 1.7 C
// > 1.3 C-
// > 1.0 D+
// > 0.7 D
// > 0.0 D-
// 0.0 E
// Example:
// >>> gradeEquation((new ArrayList<Float>(Arrays.asList((float)4.0f, (float)3l, (float)1.7f, (float)2l, (float)3.5f))))
// (new ArrayList<String>(Arrays.asList((String)"A+", (String)"B", (String)"C-", (String)"C", (String)"A-")))
public static ArrayList<String> numericalLetterGrade(ArrayList<Float> grades) {
ArrayList<String> letterGrades = new ArrayList<String>();
for (int i = 0; i < grades.size(); i++) {
float currentGrade = grades.get(i);
if (currentGrade == 4.0f) {
letterGrades.add("A+");
} else if (currentGrade > 3.7f) {
letterGrades.add("A");
} else if (currentGrade > 3.3f) {
letterGrades.add("A-");
} else if (currentGrade > 3.0f) {
letterGrades.add("B+");
} else if (currentGrade > 2.7f) {
letterGrades.add("B");
} else if (currentGrade > 2.3f) {
letterGrades.add("B-");
} else if (currentGrade > 2.0f) {
letterGrades.add("C+");
} else if (currentGrade > 1.7f) {
letterGrades.add("C");
} else if (currentGrade > 1.3f) { | letterGrades.add("C-"); | } else if (currentGrade > 1.0f) {
letterGrades.add("D+");
} else if (currentGrade > 0.7f) {
letterGrades.add("D");
} else if (currentGrade > 0.0f) {
letterGrades.add("D-");
} else {
letterGrades.add("E");
}
}
return letterGrades;
}
} |
//Given a positive integer, obtain its roman numeral equivalent as a string,
// and return it in lowercase.
// Restrictions: 1 <= num <= 1000
// Examples:
// >>> int_to_mini_roman(19)
// "xix"
// >>> int_to_mini_roman(152)
// "clii"
// >>> int_to_mini_roman(426)
// "cdxxvi"
function int_to_mini_roman(number){
var roman_conversion = {
1000: "m",
900: "cm",
500: "d",
400: "cd",
100: "c",
90: "xc",
50: "l",
40: "xl",
10: "x",
9: "ix",
5: "v",
4: "iv",
1: "i"
};
var output = "";
while (number > 0){
var key_list = Object.keys(roman_conversion).map(Number);
key_list.sort(function(a, b){return b-a});
for (var i = 0; i < key_list.length; i++){
if (key_list[i] <= number){
output += roman_conversion[key_list[i]];
number -= key_list[i];
break; | } | }
}
return output;
}
|
//Given a positive integer n, return the product of the odd digits.
// Return 0 if all digits are even.
// For example:
// >>> digits(1)
// 1
// >>> digits(4)
// 0
// >>> digits(235)
// 15
function digits(n){
if(!n){
return "error";
}
if(n < 0){
return "error";
}
var temp = 0;
var prod = 1;
var odd = true;
for(n; n > 0; n = Math.floor(n/10)){
temp = n % 10;
if(temp % 2 != 0){
prod *= temp;
odd = false;
}
}
if(odd){
return 0;
} | return prod; | }
|
//Given an array of strings, where each string consists of only digits, return an array.
// Each element i of the output should be "the number of odd elements in the
// string i of the input." where all the i's should be replaced by the number
// of odd digits in the i'th string of the input.
// >>> odd_count(["1234567"])
// ["the number of odd elements 4n the str4ng 4 of the 4nput."]
// >>> odd_count(["3", "11111111"])
// ["the number of odd elements 1n the str1ng 1 of the 1nput.", "the number of odd elements 8n the str8ng 8 of the 8nput."]
function odd_count(lst){
var new_lst = [];
for (var i = 0; i < lst.length; i++){
var word = lst[i]; | var odd_sum = 0; | for (var j = 0; j < word.length; j++){
if (word[j] % 2 == 1){
odd_sum += 1;
}
}
var odd_word = "the number of odd elements " + odd_sum + "n the str" + odd_sum + "ng " + odd_sum + " of the " + odd_sum + "nput.";
new_lst.push(odd_word);
}
return new_lst;
} |
from typing import List
def numerical_letter_grade(grades: List[float]) -> List[str]:
"""It is the last week of the semester and the teacher has to give the grades
to students. The teacher has been making her own algorithm for grading.
The only problem is, she has lost the code she used for grading.
She has given you a list of GPAs for some students and you have to write
a function that can output a list of letter grades using the following table:
GPA | Letter grade
4.0 A+
> 3.7 A
> 3.3 A-
> 3.0 B+
> 2.7 B
> 2.3 B-
> 2.0 C+
> 1.7 C
> 1.3 C-
> 1.0 D+
> 0.7 D
> 0.0 D-
0.0 E
Example:
>>> grade_equation([4.0, 3, 1.7, 2, 3.5])
['A+', 'B', 'C-', 'C', 'A-']
"""
final = []
for x in grades:
if x == 4.0:
final.append('A+')
elif x > 3.7:
final.append('A')
elif x > 3.3:
final.append('A-')
elif x > 3.0:
final.append('B+')
elif x > 2.7:
final.append('B')
elif x > 2.3:
final.append('B-')
elif x > 2.0:
final.append('C+')
elif x > 1.7: | final.append('C') | elif x > 1.3:
final.append('C-')
elif x > 1.0:
final.append('D+')
elif x > 0.7:
final.append('D')
elif x > 0.0:
final.append('D-')
else:
final.append('E')
return final
|
//Write a function count_nums which takes an array of integers and returns
// the number of elements which has a sum of digits > 0.
// If a number is negative, then its first signed digit will be negative:
// e.g. -123 has signed digits -1, 2, and 3.
// >>> count_nums([])
// 0
// >>> count_nums([-1, 11, -11])
// 1
// >>> count_nums([1, 1, 2])
// 3
function count_nums(arr){
if(!arr.length) {
return 0;
}
var count = 0;
for(var i = 0; i < arr.length; i++) {
var sum = 0;
var str = arr[i].toString();
if(str[0] === '-') {
sum += Number(str[0] + str[1]);
for(var j = 2; j < str.length; j++) {
sum += Number(str[j]);
}
} else {
for(var k = 0; k < str.length; k++) {
sum += Number(str[k]);
}
}
if(sum > 0) {
count++; | } | }
return count;
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given the lengths of the three sides of a triangle. Return the area of
// the triangle rounded to 2 decimal points if the three sides form a valid triangle.
// Otherwise return -1
// Three sides make a valid triangle when the sum of any two sides is greater
// than the third side.
// Example:
// >>> triangleArea((3l), (4l), (5l))
// (6.0f)
// >>> triangleArea((1l), (2l), (10l))
// (float)-1l
public static float triangleArea(long a, long b, long c) {
if(a + b > c && a + c > b && b + c > a) {
float s = (a + b + c) / 2.0f; | float area = (float)Math.sqrt(s * (s - a) * (s - b) * (s - c)); | return (float)Math.round(area * 100) / 100;
}
return -1;
}
} |
//You are given a non-empty array of positive integers. Return the greatest integer that is greater than
// zero, and has a frequency greater than or equal to the value of the integer itself.
// The frequency of an integer is the number of times it appears in the array.
// If no such a value exist, return -1.
// Examples:
// >>> search([4, 1, 2, 2, 3, 1])
// 2
// >>> search([1, 2, 2, 3, 3, 3, 4, 4, 4])
// 3
// >>> search([5, 5, 4, 4, 4])
// -1
function search(lst){
var h = {};
for (var i=0; i<lst.length; i++){
var x = lst[i];
if (x in h){
h[x] += 1;
}
else{
h[x] = 1;
}
}
var keys = Object.keys(h);
var max = 0; | for (var i=0; i<keys.length; i++){ | if (keys[i] > max && h[keys[i]] >= keys[i]){
max = keys[i];
}
}
if (max == 0){
return -1;
}
return max;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// brackets is a string of "<" and ">".
// return true if every opening bracket has a corresponding closing bracket.
// >>> correctBracketing(("<"))
// (false)
// >>> correctBracketing(("<>"))
// (true)
// >>> correctBracketing(("<<><>>"))
// (true)
// >>> correctBracketing(("><<>"))
// (false)
public static boolean correctBracketing(String brackets) {
int open = 0; | for (int i = 0; i < brackets.length(); i++) { | char c = brackets.charAt(i);
if (c == '<') {
open++;
} else {
open--;
}
if (open < 0) {
return false;
}
}
return open == 0;
}
} |
//Check if in given array of numbers, are any two numbers closer to each other than
// given threshold.
// >>> has_close_elements([1.0, 2.0, 3.0], 0.5)
// false
// >>> has_close_elements([1.0, 2.8, 3.0, 4.0, 5.0, 2.0], 0.3)
// true
function has_close_elements(numbers, threshold){
for(var i = 0; i < numbers.length; i++){
for(var j = i + 1; j < numbers.length; j++){ | if(Math.abs(numbers[i] - numbers[j]) < threshold){ | return true;
}
}
}
return false;
}
|
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Create a function which takes a string representing a file's name, and returns
// 'Yes' if the the file's name is valid, and returns 'No' otherwise.
// A file's name is considered to be valid if and only if all the following conditions
// are met:
// - There should not be more than three digits ('0'-'9') in the file's name.
// - The file's name contains exactly one dot '.'
// - The substring before the dot should not be empty, and it starts with a letter from
// the latin alphapet ('a'-'z' and 'A'-'Z').
// - The substring after the dot should be one of these: ['txt', 'exe', 'dll']
// Examples:
// >>> fileNameCheck(("example.txt"))
// ("Yes")
// >>> fileNameCheck(("1example.dll"))
// ("No")
public static String fileNameCheck(String file_name) {
String result = "No";
String[] tokens = file_name.split("\\.");
if (tokens.length != 2) { | return result; | }
String extension = tokens[1];
if (!(extension.equals("txt") || extension.equals("exe") || extension.equals("dll"))) {
return result;
}
String name = tokens[0];
int count = 0;
for (int i = 0; i < name.length(); i++) {
if (Character.isDigit(name.charAt(i))) {
count++;
}
}
if (count > 3) {
return result;
}
if (name.length() == 0) {
return result;
}
if (!Character.isLetter(name.charAt(0))) {
return result;
}
return "Yes";
}
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Write a function that takes a message, and encodes in such a
// way that it swaps case of all letters, replaces all vowels in
// the message with the letter that appears 2 places ahead of that
// vowel in the english alphabet.
// Assume only letters.
// Examples:
// >>> encode(("test"))
// ("TGST")
// >>> encode(("This is a message"))
// ("tHKS KS C MGSSCGG")
public static String encode(String message) {
char[] arr = message.toCharArray();
char[] vowels = "AEIOUaeiou".toCharArray();
for (int i = 0; i < arr.length; i++) {
if (Arrays.binarySearch(vowels, arr[i]) > -1) {
arr[i] = (char) (arr[i] + 2);
}
if (Character.isUpperCase(arr[i])) {
arr[i] = Character.toLowerCase(arr[i]);
} else {
arr[i] = Character.toUpperCase(arr[i]);
} | } | return new String(arr);
}
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Given a positive integer, obtain its roman numeral equivalent as a string,
// and return it in lowercase.
// Restrictions: 1 <= num <= 1000
// Examples:
// >>> intToMiniRoman((19l))
// ("xix")
// >>> intToMiniRoman((152l))
// ("clii")
// >>> intToMiniRoman((426l))
// ("cdxxvi")
public static String intToMiniRoman(long number) {
int num = (int) number;
String res = "";
if (num == 0)
return "";
while (num >= 1000) {
res += "m";
num -= 1000;
}
if (num >= 900) {
res += "cm";
num -= 900;
}
if (num >= 500) {
res += "d";
num -= 500;
}
if (num >= 400) {
res += "cd";
num -= 400;
}
while (num >= 100) {
res += "c";
num -= 100;
}
if (num >= 90) {
res += "xc";
num -= 90;
}
if (num >= 50) {
res += "l";
num -= 50;
}
if (num >= 40) {
res += "xl";
num -= 40;
}
while (num >= 10) {
res += "x";
num -= 10;
}
if (num >= 9) {
res += "ix";
num -= 9;
}
if (num >= 5) {
res += "v";
num -= 5;
}
if (num >= 4) {
res += "iv";
num -= 4;
}
while (num >= 1) { | res += "i"; | num -= 1;
}
return res;
}
} |
//You are given an array of integers.
// You need to find the largest prime value and return the sum of its digits.
// Examples:
// >>> skjkasdkd([0, 3, 2, 1, 3, 5, 7, 4, 5, 5, 5, 2, 181, 32, 4, 32, 3, 2, 32, 324, 4, 3])
// 10
// >>> skjkasdkd([1, 0, 1, 8, 2, 4597, 2, 1, 3, 40, 1, 2, 1, 2, 4, 2, 5, 1])
// 25
// >>> skjkasdkd([1, 3, 1, 32, 5107, 34, 83278, 109, 163, 23, 2323, 32, 30, 1, 9, 3])
// 13
// >>> skjkasdkd([0, 724, 32, 71, 99, 32, 6, 0, 5, 91, 83, 0, 5, 6])
// 11
// >>> skjkasdkd([0, 81, 12, 3, 1, 21])
// 3
// >>> skjkasdkd([0, 8, 1, 2, 1, 7])
// 7
function skjkasdkd(lst){
function prime(n){
if (n<2) return false;
for (let i=2; i<=Math.sqrt(n); i++){
if (n%i==0) return false;
}
return true;
}
let max = Math.max(...lst.filter(x=>prime(x)));
let digits = String(max).split('');
let sum = 0;
for (let i=0; i<digits.length; i++){ | sum += Number(digits[i]); | }
return sum;
} |
//Return true if a given number is prime, and false otherwise.
// >>> is_prime(6)
// false
// >>> is_prime(101)
// true
// >>> is_prime(11)
// true
// >>> is_prime(13441)
// true
// >>> is_prime(61)
// true
// >>> is_prime(4)
// false
// >>> is_prime(1)
// false
function is_prime(n){
if (n==1) {
return false;
}
if (n==2) {
return true;
}
| if (n%2==0) { | return false;
}
var limit = Math.sqrt(n);
for (var i = 3; i <= limit; i += 2) {
if (n%i==0) {
return false;
}
}
return true;
}
|
//In this problem, you will implement a function that takes two arrays of numbers,
// and determines whether it is possible to perform an exchange of elements
// between them to make lst1 an array of only even numbers.
// There is no limit on the number of exchanged elements between lst1 and lst2.
// If it is possible to exchange elements between the lst1 and lst2 to make
// all the elements of lst1 to be even, return "YES".
// Otherwise, return "NO".
// For example:
// >>> exchange([1, 2, 3, 4], [1, 2, 3, 4])
// "YES"
// >>> exchange([1, 2, 3, 4], [1, 5, 3, 4])
// "NO"
// It is assumed that the input arrays will be non-empty.
function exchange(lst1, lst2){
var num1 = 0;
var num2 = 0;
var exchange = true;
while(num1 < lst1.length && exchange){
if(lst1[num1] % 2 !== 0){
exchange = false;
if(num2 < lst2.length){
while(num2 < lst2.length && exchange == false){
if(lst2[num2] % 2 == 0){
lst1[num1] = lst2[num2];
exchange = true;
}
num2++;
}
}
}
num1++;
}
if(exchange){
return "YES";
}else{ | return "NO"; | }
}
|
from typing import Union
def compare_one(a: Union[int, float, str], b: Union[int, float, str]) -> Union[int, float, str, None]:
"""
Create a function that takes integers, floats, or strings representing
real numbers, and returns the larger variable in its given variable type.
Return None if the values are equal.
Note: If a real number is represented as a string, the floating point might be . or ,
>>> compare_one(1, 2.5)
2.5
>>> compare_one(1, '2,3')
'2,3'
>>> compare_one('5,1', '6')
'6'
>>> compare_one('1', 1)
None
"""
if isinstance(a, int) and isinstance(b, int):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, float) and isinstance(b, float):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, str) and isinstance(b, str):
a_tmp = a.replace(',', '.')
b_tmp = b.replace(',', '.')
if float(a_tmp) > float(b_tmp):
return a
elif float(a_tmp) < float(b_tmp):
return b
else:
return None | elif isinstance(a, int) and isinstance(b, float): | if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, float) and isinstance(b, int):
if a > b:
return a
elif a < b:
return b
else:
return None
elif isinstance(a, str) and isinstance(b, int):
a_tmp = a.replace(',', '.')
if float(a_tmp) > b:
return a
elif float(a_tmp) < b:
return b
else:
return None
elif isinstance(a, int) and isinstance(b, str):
b_tmp = b.replace(',', '.')
if a > float(b_tmp):
return a
elif a < float(b_tmp):
return b
else:
return None
elif isinstance(a, str) and isinstance(b, float):
a_tmp = a.replace(',', '.')
if float(a_tmp) > b:
return a
elif float(a_tmp) < b:
return b |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// You are given an array array list of integers.
// You need to find the largest prime value and return the sum of its digits.
// Examples:
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)3l, (long)2l, (long)1l, (long)3l, (long)5l, (long)7l, (long)4l, (long)5l, (long)5l, (long)5l, (long)2l, (long)181l, (long)32l, (long)4l, (long)32l, (long)3l, (long)2l, (long)32l, (long)324l, (long)4l, (long)3l))))
// (10l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)1l, (long)0l, (long)1l, (long)8l, (long)2l, (long)4597l, (long)2l, (long)1l, (long)3l, (long)40l, (long)1l, (long)2l, (long)1l, (long)2l, (long)4l, (long)2l, (long)5l, (long)1l))))
// (25l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)1l, (long)32l, (long)5107l, (long)34l, (long)83278l, (long)109l, (long)163l, (long)23l, (long)2323l, (long)32l, (long)30l, (long)1l, (long)9l, (long)3l))))
// (13l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)724l, (long)32l, (long)71l, (long)99l, (long)32l, (long)6l, (long)0l, (long)5l, (long)91l, (long)83l, (long)0l, (long)5l, (long)6l))))
// (11l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)81l, (long)12l, (long)3l, (long)1l, (long)21l))))
// (3l)
// >>> skjkasdkd((new ArrayList<Long>(Arrays.asList((long)0l, (long)8l, (long)1l, (long)2l, (long)1l, (long)7l))))
// (7l)
public static long skjkasdkd(ArrayList<Long> lst) {
long largestPrime = 0l;
for (Long item : lst) {
long number = item;
int counter = 0;
if (item == 1) {
continue;
}
if (item == 2) {
counter = 1;
}
else {
int l = 1;
while (l <= (int) number) {
if (number % l == 0) { | counter++; | }
l++;
}
}
if (counter == 2) {
if (number > largestPrime) {
largestPrime = number;
}
}
}
if (largestPrime == 0) {
return 0l;
}
else {
long output = 0l;
while (largestPrime > 0) {
output += largestPrime % 10;
largestPrime /= 10;
}
return output;
}
}
} |
import java.util.*;
import java.lang.reflect.*;
import org.javatuples.*;
import java.security.*;
import java.math.*;
import java.io.*;
import java.util.stream.*;
class Problem {
// Evaluate whether the given number n can be written as the sum of exactly 4 positive even numbers
// Example
// >>> isEqualToSumEven((4l))
// (false)
// >>> isEqualToSumEven((6l))
// (false)
// >>> isEqualToSumEven((8l))
// (true)
public static boolean isEqualToSumEven(long n) {
if (n < 4) {
return false;
}
long d = 2;
if (n % 2 == 1) {
d = 1;
}
long minX = d;
long maxX = n / 4;
while (minX <= maxX) {
long x = (minX + maxX) / 2;
long y = n - 2 * x;
if (y % 2 == 1) {
y--; | } | if (2 * x + y < n) {
minX = x + 1;
} else if (2 * x + y > n) {
maxX = x - 1;
} else {
return true;
}
}
return false;
}
} |
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