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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;
} |
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 returns true if the given number is the multiplication of 3 prime numbers
// and false otherwise.
// Knowing that (a) is less then 100.
// Example:
// >>> isMultiplyPrime((30l))
// (true)
// 30 = 2 * 3 * 5
public static boolean isMultiplyPrime(long a) {
if (a == 1) {
return false;
}
if (a == 2) {
return true;
} | long copyOfInput = a; | ArrayList<Long> arr = new ArrayList<Long>();
while (a % 2 == 0) {
arr.add(2l);
a /= 2;
}
for (long i = 3; i <= Math.sqrt(a); i += 2) {
while (a % i == 0) {
arr.add(i);
a /= i;
}
}
if (a > 2) {
arr.add(a);
}
return arr.size() == 3;
}
} |
from typing import List
def max_fill(grid: List[List[int]], capacity: int) -> int:
"""
You are given a rectangular grid of wells. Each row represents a single well,
and each 1 in a row represents a single unit of water.
Each well has a corresponding bucket that can be used to extract water from it,
and all buckets have the same capacity.
Your task is to use the buckets to empty the wells.
Output the number of times you need to lower the buckets.
Example 1:
>>> max_fill([[0, 0, 1, 0], [0, 1, 0, 0], [1, 1, 1, 1]], 1)
6
Example 2:
>>> max_fill([[0, 0, 1, 1], [0, 0, 0, 0], [1, 1, 1, 1], [0, 1, 1, 1]], 2)
5
Example 3:
>>> max_fill([[0, 0, 0], [0, 0, 0]], 5)
0
Constraints:
* all wells have the same length
* 1 <= grid.length <= 10^2
* 1 <= grid[:,1].length <= 10^2
* grid[i][j] -> 0 | 1
* 1 <= capacity <= 10
"""
times = 0
for row in grid: | for i in range(0, len(row), capacity): | if 1 in row[i:i+capacity]:
times += 1
return times
|
//You are given a string representing a sentence,
// the sentence contains some words separated by a space,
// and you have to return a string that contains the words from the original sentence,
// whose lengths are prime numbers,
// the order of the words in the new string should be the same as the original one.
// Example 1:
// >>> words_in_sentence("This is a test")
// "is"
// Example 2:
// >>> words_in_sentence("lets go for swimming")
// "go for"
// Constraints:
// * 1 <= len(sentence) <= 100
// * sentence contains only letters
function words_in_sentence(sentence){
return sentence.split(' ').filter(word => {
var number = word.length;
var isPrime = true; | if(number === 1 || number === 0){ | return false;
}
for(var i = 2; i < number; i++){
if(number % i === 0){
isPrime = false;
break;
}
}
return isPrime;
}).join(' ');
} |
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 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:
// >>> checkIfLastCharIsALetter(("apple pie"))
// (false)
// >>> checkIfLastCharIsALetter(("apple pi e"))
// (true)
// >>> checkIfLastCharIsALetter(("apple pi e "))
// (false)
// >>> checkIfLastCharIsALetter((""))
// (false)
public static boolean checkIfLastCharIsALetter(String txt) {
if (txt.length() == 0) {
return false;
}
int len = txt.length();
if (txt.substring(len - 1, len).matches("[A-Z]") || txt.substring(len - 1, len).matches("[a-z]")) {
if (len == 1) {
return true;
}
if (txt.substring(len - 2, len - 1).matches(" ")) {
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 {
// 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 balance = 0;
for (int i = 0; i < brackets.length(); i++) { | if (brackets.charAt(i) == '(') { | balance++;
} else {
balance--;
}
if (balance < 0) {
return false;
}
}
return balance == 0;
}
} |
//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 of GPAs for some students and you have to write
// a function that can output an array 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-"]
function numerical_letter_grade(grades){
var grades_array = [];
for (var i = 0; i < grades.length; i++){
if (grades[i] >= 4.0){
grades_array.push("A+");
}
else if (grades[i] > 3.7){
grades_array.push("A");
}
else if (grades[i] > 3.3){
grades_array.push("A-");
}
else if (grades[i] > 3.0){
grades_array.push("B+");
}
else if (grades[i] > 2.7){
grades_array.push("B");
}
else if (grades[i] > 2.3){
grades_array.push("B-");
}
else if (grades[i] > 2.0){
grades_array.push("C+");
}
else if (grades[i] > 1.7){ | grades_array.push("C"); | }
else if (grades[i] > 1.3){
grades_array.push("C-");
}
else if (grades[i] > 1.0){
grades_array.push("D+");
}
else if (grades[i] > 0.7){
grades_array.push("D");
}
else if (grades[i] > 0.0){
grades_array.push("D-");
}
else if (grades[i] == 0.0){
grades_array.push("E");
}
}
return grades_array;
} |
def encode(message: str) -> str:
"""
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'
"""
def swap_case(ch: str) -> str: | if ch.isupper(): | return ch.lower()
else:
return ch.upper()
def encode_vowel(ch: str) -> str:
if ch.lower() in 'aeiou':
if ch.lower() in 'wxyz':
return ch.lower()
return chr(ord(ch) + 2)
return ch
return ''.join(map(lambda ch: encode_vowel(swap_case(ch)), message))
|
//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 of GPAs for some students and you have to write
// a function that can output an array 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-"]
function numerical_letter_grade(grades){
var grades_array = [];
for (var i = 0; i < grades.length; i++){
if (grades[i] >= 4.0){
grades_array.push("A+");
}
else if (grades[i] > 3.7){ | grades_array.push("A"); | }
else if (grades[i] > 3.3){
grades_array.push("A-");
}
else if (grades[i] > 3.0){
grades_array.push("B+");
}
else if (grades[i] > 2.7){
grades_array.push("B");
}
else if (grades[i] > 2.3){
grades_array.push("B-");
}
else if (grades[i] > 2.0){
grades_array.push("C+");
}
else if (grades[i] > 1.7){
grades_array.push("C");
}
else if (grades[i] > 1.3){
grades_array.push("C-");
}
else if (grades[i] > 1.0){
grades_array.push("D+");
}
else if (grades[i] > 0.7){
grades_array.push("D");
}
else if (grades[i] > 0.0){
grades_array.push("D-");
}
else if (grades[i] == 0.0){
grades_array.push("E");
}
}
return grades_array;
} |
//You are given a 2 dimensional data, as a nested arrays,
// which is similar to matrix, however, unlike matrices,
// each row may contain a different number of columns.
// Given lst, and integer x, find integers x in the array,
// and return array of arrays, [(x1, y1), (x2, y2) ...] such that
// each array is a coordinate - (row, columns), starting with 0.
// Sort coordinates initially by rows in ascending order.
// Also, sort coordinates of the row by columns in descending order.
// Examples:
// >>> get_row([[1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 1, 6], [1, 2, 3, 4, 5, 1]], 1)
// [[0, 0], [1, 4], [1, 0], [2, 5], [2, 0]]
// >>> get_row([], 1)
// []
// >>> get_row([[], [1], [1, 2, 3]], 3)
// [[2, 2]]
function get_row(lst, x){
let coordinates = [];
for (let i = 0; i < lst.length; i++){
let array = lst[i];
for (let j = 0; j < array.length; j++){
if (lst[i][j] === x){
coordinates.push([i, j]);
}
}
}
return coordinates.sort(function(a, b){ | return a[0] - b[0] || b[1] - a[1]; | });
}
|
//You are given a 2 dimensional data, as a nested arrays,
// which is similar to matrix, however, unlike matrices,
// each row may contain a different number of columns.
// Given lst, and integer x, find integers x in the array,
// and return array of arrays, [(x1, y1), (x2, y2) ...] such that
// each array is a coordinate - (row, columns), starting with 0.
// Sort coordinates initially by rows in ascending order.
// Also, sort coordinates of the row by columns in descending order.
// Examples:
// >>> get_row([[1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 1, 6], [1, 2, 3, 4, 5, 1]], 1)
// [[0, 0], [1, 4], [1, 0], [2, 5], [2, 0]]
// >>> get_row([], 1)
// []
// >>> get_row([[], [1], [1, 2, 3]], 3)
// [[2, 2]]
function get_row(lst, x){
let coordinates = []; | for (let i = 0; i < lst.length; i++){ | let array = lst[i];
for (let j = 0; j < array.length; j++){
if (lst[i][j] === x){
coordinates.push([i, j]);
}
}
}
return coordinates.sort(function(a, b){
return a[0] - b[0] || b[1] - a[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 {
// You are given a string s.
// if s[i] is a letter, reverse its case from lower to upper or vise versa,
// otherwise keep it as it is.
// If the string contains no letters, reverse the string.
// The function should return the resulted string.
// Examples
// >>> solve(("1234"))
// ("4321")
// >>> solve(("ab"))
// ("AB")
// >>> solve(("#a@C"))
// ("#A@c")
public static String solve(String s) {
boolean letterNotFound = true;
StringBuilder sb = new StringBuilder();
if (s.length() == 0) return s; | for (int i = 0; i < s.length(); i++) { | if (Character.isLetter(s.charAt(i))) {
letterNotFound = false;
if (Character.isUpperCase(s.charAt(i))) {
sb.append(Character.toLowerCase(s.charAt(i)));
} else {
sb.append(Character.toUpperCase(s.charAt(i)));
}
} else {
sb.append(s.charAt(i));
}
}
if (letterNotFound) {
return sb.reverse().toString();
}
return sb.toString();
}
} |
//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;
}
|
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"; | }
} |
//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";
}
}
|
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 n, return a sorted array list 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 list 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.
// >>> getOddCollatz((5l))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)5l)))
public static ArrayList<Long> getOddCollatz(long n) {
if (n <= 0) return new ArrayList<Long>();
HashSet<Long> odds = new HashSet<Long>();
long n1 = n; | while (n1 > 1) { | if (n1 % 2 == 1) odds.add(n1);
n1 = n1 % 2 == 0 ? n1 / 2 : 3 * n1 + 1;
}
odds.add(1l);
Long[] odds_arr = odds.toArray(new Long[odds.size()]);
ArrayList<Long> ans = new ArrayList<Long>(Arrays.asList(odds_arr));
Collections.sort(ans);
return ans;
}
} |
//Return sorted unique elements in an array
// >>> unique([5, 3, 5, 2, 3, 3, 9, 0, 123])
// [0, 2, 3, 5, 9, 123]
function unique(l){
return l.filter(function(e, i, a){ | return a.indexOf(e) === i; | }).sort(function(a, b){return a-b});
}
|
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
|
from typing import List
def by_length(arr: List[int]) -> List[str]:
"""
Given an array of integers, sort the integers that are between 1 and 9 inclusive,
reverse the resulting array, and then replace each digit by its corresponding name from
"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine".
For example:
>>> by_length([2, 1, 1, 4, 5, 8, 2, 3])
['Eight', 'Five', 'Four', 'Three', 'Two', 'Two', 'One', 'One']
If the array is empty, return an empty array:
>>> by_length([])
[]
If the array has any strange number ignore it:
>>> by_length([1, -1, 55])
['One']
"""
digits = [x for x in arr if 1 <= x <= 9]
digits.sort() | digits.reverse() | for i in range(len(digits)):
digits[i] = ['One', 'Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine'][digits[i] - 1]
return digits
|
from typing import List
def search(lst: List[int]) -> int:
"""
You are given a non-empty list 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 list.
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
"""
dic = dict()
for i in lst:
if i in dic:
dic[i] += 1
else: | dic[i] = 1 | maxValue = -1
for key, value in dic.items():
if value >= key and key > maxValue:
maxValue = key
return maxValue
|
//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;
}
|
from typing import List
def do_algebra(operator: List[str], operand: List[int]) -> int:
"""
Given two lists operator, and operand. The first list has basic algebra operations, and
the second list is a list of integers. Use the two given 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 = [2, 3, 4, 5]
result = 2 + 3 * 4 - 5
=> result = 9
Note:
The length of operator list is equal to the length of operand list minus one.
Operand is a list of of non-negative integers.
Operator list has at least one operator, and operand list has at least two operands.
"""
if len(operator) != len(operand) - 1:
raise ValueError("Invalid number of operators")
expr = [str(operand[0])]
for idx, op in enumerate(operator):
expr.append(op)
expr.append(str(operand[idx + 1])) | return eval("".join(expr)) | |
//Write a function that returns true if the object q will fly, and false otherwise.
// The object q will fly if it's balanced (it is a palindromic array) and the sum of its elements is less than or equal the maximum possible weight w.
// Example:
// >>> will_it_fly([1, 2], 5)
// false
// # 1+2 is less than the maximum possible weight, but it's unbalanced.
// >>> will_it_fly([3, 2, 3], 1)
// false
// # it's balanced, but 3+2+3 is more than the maximum possible weight.
// >>> will_it_fly([3, 2, 3], 9)
// true
// # 3+2+3 is less than the maximum possible weight, and it's balanced.
// >>> will_it_fly([3], 5)
// true
// # 3 is less than the maximum possible weight, and it's balanced.
function will_it_fly(q, w){
let sum = 0;
let balanced = true;
for(let i = 0; i < q.length; i++){
sum += q[i];
}
for(let i = 0; i < q.length; i++){
if(q[i] !== q[q.length - 1 - i]){
balanced = false;
break;
}
} | return balanced && sum <= w; | } |
//The FibFib number sequence is a sequence similar to the Fibbonacci sequnece that's defined as follows:
// fibfib(0) == 0
// fibfib(1) == 0
// fibfib(2) == 1
// fibfib(n) == fibfib(n-1) + fibfib(n-2) + fibfib(n-3).
// Please write a function to efficiently compute the n-th element of the fibfib number sequence.
// >>> fibfib(1)
// 0
// >>> fibfib(5)
// 4
// >>> fibfib(8)
// 24
function fibfib(n){
if(n <= 0){
return 0;
}
if(n == 1){
return 0;
} | if(n == 2){ | return 1;
}
return fibfib(n-1) + fibfib(n-2) + fibfib(n-3);
}
|
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 two strings, both strings consist of open
// parentheses '(' or close parentheses ')' only.
// Your job is to check if it is possible to concatenate the two strings in
// some order, that the resulting string will be good.
// A string S is considered to be good if and only if all parentheses in S
// are balanced. For example: the string '(())()' is good, while the string
// '())' is not.
// Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.
// Examples:
// >>> matchParens((new ArrayList<String>(Arrays.asList((String)"()(", (String)")"))))
// ("Yes")
// >>> matchParens((new ArrayList<String>(Arrays.asList((String)")", (String)")"))))
// ("No")
public static String matchParens(ArrayList<String> lst) {
String s1 = lst.get(0);
String s2 = lst.get(1);
String s3 = s1+s2;
String s4 = s2+s1;
int c1 = 0;
int c2 = 0;
for (int i = 0; i < s3.length(); i++) {
if (s3.charAt(i) == '(') c1 += 1;
else c1 -= 1;
if (c1 < 0) break;
}
for (int i = 0; i < s4.length(); i++) {
if (s4.charAt(i) == '(') c2 += 1; | else c2 -= 1; | if (c2 < 0) break;
}
if (c1 == 0 || c2 == 0) return "Yes";
else return "No";
}
} |
//"Given an array representing a branch of a tree that has non-negative integer nodes
// your task is to pluck one of the nodes and return it.
// The plucked node should be the node with the smallest even value.
// If multiple nodes with the same smallest even value are found return the node that has smallest index.
// The plucked node should be returned in an array, [ smalest_value, its index ],
// If there are no even values or the given array is empty, return [].
// Example 1:
// >>> pluck([4, 2, 3])
// [2, 1]
// Explanation: 2 has the smallest even value, and 2 has the smallest index.
// Example 2:
// >>> pluck([1, 2, 3])
// [2, 1]
// Explanation: 2 has the smallest even value, and 2 has the smallest index.
// Example 3:
// >>> pluck([])
// []
// Example 4:
// >>> pluck([5, 0, 3, 0, 4, 2])
// [0, 1]
// Explanation: 0 is the smallest value, but there are two zeros,
// so we will choose the first zero, which has the smallest index.
// Constraints:
// * 1 <= nodes.length <= 10000
// * 0 <= node.value
function pluck(arr){
let result = [],
index = -1;
for(let i = 0; i < arr.length; i++){
if(arr[i] % 2 === 0){
if(result.length === 0 || result[0] > arr[i]){
result = [arr[i], i];
}
}
} | return result; | }
|
//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 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;
}
|
//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;
} |
from typing import List
def double_the_difference(lst: List[float]) -> int:
"""
Given a list of numbers, return the sum of squares of the numbers
in the list that are odd. Ignore numbers that are negative or not integers.
>>> double_the_difference([1, 3, 2, 0])
10
>>> double_the_difference([-1, -2, 0])
0
>>> double_the_difference([9, -2])
81
>>> double_the_difference([0])
0
If the input list is empty, return 0.
"""
sum = 0
for x in lst:
if not x < 0 and x % 2 == 1: | sum += x * x | return sum
|
//In this Kata, you have to sort an array of non-negative integers according to
// number of ones in their binary representation in ascending order.
// For similar number of ones, sort based on decimal value.
// It must be implemented like this:
// >>> sort_array([1, 5, 2, 3, 4])
// [1, 2, 3, 4, 5]
// >>> sort_array([-2, -3, -4, -5, -6])
// [-6, -5, -4, -3, -2]
// >>> sort_array([1, 0, 2, 3, 4])
// [0, 1, 2, 3, 4]
function sort_array(arr){
return arr.sort((a, b) => {
let aBinary = a.toString(2).split('').filter(el => el === '1').length;
let bBinary = b.toString(2).split('').filter(el => el === '1').length;
if (aBinary === bBinary) {
return a - b;
} else {
return aBinary - bBinary;
} | }); | } |
//You will be given the name of a class (a string) and an array of extensions.
// The extensions are to be used to load additional classes to the class. The
// strength of the extension is as follows: Let CAP be the number of the uppercase
// letters in the extension's name, and let SM be the number of lowercase letters
// in the extension's name, the strength is given by the fraction CAP - SM.
// You should find the strongest extension and return a string in this
// format: ClassName.StrongestExtensionName.
// If there are two or more extensions with the same strength, you should
// choose the one that comes first in the array.
// For example, if you are given "Slices" as the class and an array of the
// extensions: ['SErviNGSliCes', 'Cheese', 'StuFfed'] then you should
// return 'Slices.SErviNGSliCes' since 'SErviNGSliCes' is the strongest extension
// (its strength is -1).
// Example:
// >>> Strongest_Extension("my_class", ["AA", "Be", "CC"])
// "my_class.AA"
function Strongest_Extension(class_name, extensions){
var output = class_name + "."; | var min_strength = 1; | for (var i = 0; i < extensions.length; i++) {
var strength = -1*(extensions[i].replace(/[^A-Z]/g, "").length - extensions[i].replace(/[^a-z]/g, "").length);
if (strength < min_strength) {
min_strength = strength;
output = class_name + "." + extensions[i];
}
}
return output;
}
|
def anti_shuffle(s: str) -> str:
"""
Write a function that takes a string and returns an ordered version of it.
Ordered version of string, is a string where all words (separated by space)
are replaced by a new word where all the characters arranged in
ascending order based on ascii value.
Note: You should keep the order of words and blank spaces in the sentence.
For example:
>>> anti_shuffle('Hi')
'Hi'
>>> anti_shuffle('hello')
'ehllo'
>>> anti_shuffle('Hello World!!!')
'Hello !!!Wdlor'
"""
words = s.split(' ') | shuffled_words = [''.join(sorted(word)) for word in words] | return ' '.join(shuffled_words)
|
//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;
} |
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 s, count the number of uppercase vowels in even indices.
// For example:
// >>> countUpper(("aBCdEf"))
// (1l)
// >>> countUpper(("abcdefg"))
// (0l)
// >>> countUpper(("dBBE"))
// (0l)
public static long countUpper(String s) {
return (long) IntStream.range(0, s.length())
.filter(i -> i % 2 == 0)
.filter(i -> s.charAt(i) == 'A' || s.charAt(i) == 'E' || s.charAt(i) == 'I' || s.charAt(i) == 'O' || s.charAt(i) == 'U')
.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 numbers, return whether or not they are sorted
// in ascending order. If array list has more than 1 duplicate of the same
// number, return false. Assume no negative numbers and only integers.
// Examples
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)5l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)2l, (long)4l, (long)5l))))
// (false)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l, (long)6l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l, (long)6l, (long)7l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)2l, (long)4l, (long)5l, (long)6l, (long)7l))))
// (false)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)2l, (long)3l, (long)3l, (long)4l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)2l, (long)2l, (long)3l, (long)4l))))
// (false)
public static boolean isSorted(ArrayList<Long> lst) {
if (lst.size() == 0) {
return true;
}
else {
int running_max = 0;
int previous_number = 0;
boolean first_time = true;
boolean seen_2_in_a_row = false;
for (Long number : lst) {
if (first_time) {
first_time = false;
previous_number = number.intValue();
running_max = number.intValue();
}
else {
if (running_max > number.intValue()) {
return false;
}
else {
if (previous_number == number.intValue()) {
if (seen_2_in_a_row) {
return false;
}
else {
seen_2_in_a_row = true; | } | }
else {
seen_2_in_a_row = false;
}
previous_number = number.intValue();
running_max = number.intValue();
}
}
}
return true;
}
}
} |
//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);
}
}
}
|
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 {
// remove_vowels is a function that takes string and returns string without vowels.
// >>> removeVowels((""))
// ("")
// >>> removeVowels(("abcdef"))
// ("bcdf")
// >>> removeVowels(("aaaaa"))
// ("")
// >>> removeVowels(("aaBAA"))
// ("B")
// >>> removeVowels(("zbcd"))
// ("zbcd")
public static String removeVowels(String text) {
String vowels = "aeiouAEIOU";
String newText = "";
for (int i = 0; i < text.length(); i++) {
if (vowels.indexOf(text.charAt(i)) == -1) {
newText += text.charAt(i);
}
} | return newText; | }
} |
//brackets is a string of "<" and ">".
// return true if every opening bracket has a corresponding closing bracket.
// >>> correct_bracketing("<")
// false
// >>> correct_bracketing("<>")
// true
// >>> correct_bracketing("<<><>>")
// true
// >>> correct_bracketing("><<>")
// false
function correct_bracketing(brackets){
var opens = 0;
for (var i = 0; i < brackets.length; i++) {
if (brackets[i] === "<") {
opens++;
}
else if (brackets[i] === ">") {
opens--;
}
if (opens < 0) {
return false;
}
} | return opens === 0; | }
|
//From a supplied array of numbers (of length at least two) select and return two that are the closest to each
// other and return them in order (smaller number, larger number).
// >>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.2])
// [2.0, 2.2]
// >>> find_closest_elements([1.0, 2.0, 3.0, 4.0, 5.0, 2.0])
// [2.0, 2.0]
function find_closest_elements(numbers){
let closest = Math.abs(numbers[0] - numbers[1]);
let a = numbers[0];
let b = numbers[1];
for (let i = 0; i < numbers.length; i++) {
for (let j = i + 1; j < numbers.length; j++) {
if (Math.abs(numbers[i] - numbers[j]) < closest) {
closest = Math.abs(numbers[i] - numbers[j]); | a = numbers[i]; | b = numbers[j];
}
}
}
return a > b ? [b, a] : [a, b];
}
|
//Given a non-empty array of integers arr and an integer k, return
// the sum of the elements with at most two digits from the first k elements of arr.
// Example:
// >>> add_elements([111, 21, 3, 4000, 5, 6, 7, 8, 9], 4)
// 24
// Constraints:
// 1. 1 <= len(arr) <= 100
// 2. 1 <= k <= len(arr)
function add_elements(arr, k){
var sum = 0;
for(var i = 0; i < k; i++){
if(arr[i] < 100){ | sum += arr[i]; | }
}
return sum;
}
|
//Insert a number 'delimeter' between every two consecutive elements of input array `numbers'
// >>> intersperse([], 4)
// []
// >>> intersperse([1, 2, 3], 4)
// [1, 4, 2, 4, 3]
function intersperse(numbers, delimeter){
var result = [];
for(var i = 0; i < numbers.length; i++){ | result.push(numbers[i]); | if(i !== numbers.length - 1){
result.push(delimeter);
}
}
return result;
}
|
//You are given a word. Your task is to find the closest vowel that stands between
// two consonants from the right side of the word (case sensitive).
// Vowels in the beginning and ending doesn't count. Return empty string if you didn't
// find any vowel met the above condition.
// You may assume that the given string contains English letter only.
// Example:
// >>> get_closest_vowel("yogurt")
// "u"
// >>> get_closest_vowel("FULL")
// "U"
// >>> get_closest_vowel("quick")
// ""
// >>> get_closest_vowel("ab")
// ""
function get_closest_vowel(word){
let vowels = ["a", "e", "i", "o", "u"];
let i = word.length - 1;
let result = [];
if (word.length === 1) {
return ""
} else {
while (i >= 0) {
if (vowels.includes(word[i])) {
if (i > 0 && i < word.length - 1) {
if (!vowels.includes(word[i-1]) && !vowels.includes(word[i+1])) {
result.push(word[i]);
}
}
}
i--;
}
if (result.length === 0) {
return ""
} else {
return result[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 {
// Write a function that takes an array array list of numbers as input and returns
// the number of elements in the array array list that are greater than 10 and both
// first and last digits of a number are odd (1, 3, 5, 7, 9).
// For example:
// >>> specialFilter((new ArrayList<Long>(Arrays.asList((long)15l, (long)-73l, (long)14l, (long)-15l))))
// (1l)
// >>> specialFilter((new ArrayList<Long>(Arrays.asList((long)33l, (long)-2l, (long)-3l, (long)45l, (long)21l, (long)109l))))
// (2l)
public static long specialFilter(ArrayList<Long> nums) {
class helper {
public long getFirstDigit(long n) {
long[] digits = getDigits(n);
return digits[0];
}
public long getLastDigit(long n) {
long[] digits = getDigits(n);
return digits[digits.length - 1];
}
public long[] getDigits(long n) {
ArrayList<Long> digits = new ArrayList<Long>();
while (n != 0) {
digits.add(n % 10); | n /= 10; | }
Collections.reverse(digits);
return digits.stream().mapToLong(i -> i).toArray();
}
}
helper h = new helper();
return nums.stream().filter(x -> x > 10).filter(x -> {
long firstDigit = h.getFirstDigit(x);
long lastDigit = h.getLastDigit(x);
return firstDigit % 2 != 0 && lastDigit % 2 != 0;
}).count();
}
} |
from typing import List, Optional
def next_smallest(lst: List[int]) -> Optional[int]:
"""
You are given a list of integers.
Write a function next_smallest() that returns the 2nd smallest element of the list.
Return None if there is no such element.
>>> next_smallest([1, 2, 3, 4, 5])
2
>>> next_smallest([5, 1, 4, 3, 2])
2
>>> next_smallest([])
None
>>> next_smallest([1, 1])
None
"""
if len(lst) < 2:
return None
min1 = min(lst[0], lst[1])
min2 = max(lst[0], lst[1])
for i in range(2, len(lst)):
if lst[i] < min1:
min2 = min1
min1 = lst[i]
elif lst[i] < min2:
min2 = lst[i] | return min2 if min1 != min2 else None | |
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
|
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;
}
} |
//You are given a positive integer n. You have to create an integer array 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 :
// >>> get_max_triples(5)
// 1
// Explanation:
// a = [1, 3, 7, 13, 21]
// The only valid triple is (1, 7, 13).
function get_max_triples(n){
let result = 0; | let arr = []; | for(let i = 1; i<=n; i++){
arr.push(i*i-i+1);
}
for(let i = 0; i<arr.length; i++){
for(let j = i+1; j<arr.length; j++){
for(let k = j+1; k<arr.length; k++){
if((arr[i]+arr[j]+arr[k])%3 === 0){
result++;
}
}
}
}
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 {
// Check if in given array list of numbers, are any two numbers closer to each other than
// given threshold.
// >>> hasCloseElements((new ArrayList<Float>(Arrays.asList((float)1.0f, (float)2.0f, (float)3.0f))), (0.5f))
// (false)
// >>> hasCloseElements((new ArrayList<Float>(Arrays.asList((float)1.0f, (float)2.8f, (float)3.0f, (float)4.0f, (float)5.0f, (float)2.0f))), (0.3f))
// (true)
public static boolean hasCloseElements(ArrayList<Float> numbers, float threshold) {
Collections.sort(numbers); | for (int i = 0; i < numbers.size() - 1; i++) { | if (numbers.get(i+1) - numbers.get(i) < 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 {
// 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 list 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
// >>> parseMusic(("o o| .| o| o| .| .| .| .| o o"))
// (new ArrayList<Long>(Arrays.asList((long)4l, (long)2l, (long)1l, (long)2l, (long)2l, (long)1l, (long)1l, (long)1l, (long)1l, (long)4l, (long)4l)))
public static ArrayList<Long> parseMusic(String music_string) {
| ArrayList<Long> res = new ArrayList<Long>(); | char[] chars = music_string.toCharArray();
for (int i = 0; i < chars.length; i++) {
if (chars[i] == 'o') {
if (i + 1 < chars.length && chars[i + 1] == '|') {
res.add((long)2l);
i += 1;
} else {
res.add((long)4l);
}
} else if (chars[i] == '.') {
if (i + 1 < chars.length && chars[i + 1] == '|') {
res.add((long)1l);
i += 1;
}
}
}
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 {
// Filter given array list of any javathon values only for integers
// >>> filterIntegers((new ArrayList<Object>(Arrays.asList((String)"a", (String)3.14f, (String)5l))))
// (new ArrayList<Long>(Arrays.asList((long)5l)))
// >>> filterIntegers((new ArrayList<Object>(Arrays.asList(1l, 2l, 3l, "abc", new HashMap<Long,Long>(Map.of()), new ArrayList<Long>(Arrays.asList())))))
// (new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l)))
public static ArrayList<Long> filterIntegers(ArrayList<Object> values) {
return new ArrayList<Long>(Arrays.asList(values.stream().filter(a -> a instanceof Long).map(a -> (Long)a).toArray(Long[]::new))); | } | } |
//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 {
// Given an array array list of numbers, return whether or not they are sorted
// in ascending order. If array list has more than 1 duplicate of the same
// number, return false. Assume no negative numbers and only integers.
// Examples
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)5l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)2l, (long)4l, (long)5l))))
// (false)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l, (long)6l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l, (long)6l, (long)7l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)2l, (long)4l, (long)5l, (long)6l, (long)7l))))
// (false)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)2l, (long)3l, (long)3l, (long)4l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)2l, (long)2l, (long)3l, (long)4l))))
// (false)
public static boolean isSorted(ArrayList<Long> lst) {
if (lst.size() == 0) {
return true;
}
else {
int running_max = 0;
int previous_number = 0;
boolean first_time = true;
boolean seen_2_in_a_row = false;
for (Long number : lst) {
if (first_time) {
first_time = false;
previous_number = number.intValue();
running_max = number.intValue();
}
else {
if (running_max > number.intValue()) {
return false;
}
else {
if (previous_number == number.intValue()) {
if (seen_2_in_a_row) {
return false;
}
else {
seen_2_in_a_row = true;
}
}
else {
seen_2_in_a_row = false;
}
previous_number = number.intValue();
running_max = number.intValue();
}
}
}
return true;
}
} | } | |
//The Brazilian factorial is defined as:
// brazilian_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!
// where n > 0
// For example:
// >>> special_factorial(4)
// 288
// The function will receive an integer as input and should return the special
// factorial of this integer.
function special_factorial(n){
if (n > 0) {
let fact = 1;
for (let i = 1; i <= n; i++) {
fact *= i;
} | return fact * special_factorial(n - 1); | } else {
return 1;
}
}
|
def largest_divisor(n: int) -> int:
""" For a given number n, find the largest number that divides n evenly, smaller than n
>>> largest_divisor(15)
5
"""
| divisor = n - 1 | while n % divisor != 0:
divisor -= 1
return divisor
|
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 the number of times the digit 7 appears in integers less than n which are divisible by 11 or 13.
// >>> fizzBuzz((50l))
// (0l)
// >>> fizzBuzz((78l))
// (2l)
// >>> fizzBuzz((79l))
// (3l)
public static long fizzBuzz(long n) {
long res = 0;
for (long i = 0; i < n; i++) {
if (i % 11 == 0 || i % 13 == 0) { | String s = String.valueOf(i); | for (int j = 0; j < s.length(); j++) {
if (s.charAt(j) == '7') {
res++;
}
}
}
}
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 {
// "
// This function will take an array array list of integers. For all entries in the array list, the function shall square the integer entry if its index is a
// multiple of 3 and will cube the integer entry if its index is a multiple of 4 and not a multiple of 3. The function will not
// change the entries in the array list whose indexes are not a multiple of 3 or 4. The function shall then return the sum of all entries.
// Examples:
// >>> lst
// (long)new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l))
// >>> lst
// (long)new ArrayList<Long>(Arrays.asList())
// >>> lst
// (long)new ArrayList<Long>(Arrays.asList((long)-1l, (long)-5l, (long)2l, (long)-1l, (long)-5l))
public static long sumSquares(ArrayList<Long> lst) {
int sz = lst.size();
for (int i = 0; i < sz; i++) {
if (i % 3 == 0) {
lst.set(i, lst.get(i) * lst.get(i));
} else if (i % 4 == 0) {
lst.set(i, lst.get(i) * lst.get(i) * lst.get(i));
}
}
long ans = 0;
for (int i = 0; i < sz; i++) {
ans += lst.get(i);
}
return ans;
} | } | |
//You will be given the name of a class (a string) and an array of extensions.
// The extensions are to be used to load additional classes to the class. The
// strength of the extension is as follows: Let CAP be the number of the uppercase
// letters in the extension's name, and let SM be the number of lowercase letters
// in the extension's name, the strength is given by the fraction CAP - SM.
// You should find the strongest extension and return a string in this
// format: ClassName.StrongestExtensionName.
// If there are two or more extensions with the same strength, you should
// choose the one that comes first in the array.
// For example, if you are given "Slices" as the class and an array of the
// extensions: ['SErviNGSliCes', 'Cheese', 'StuFfed'] then you should
// return 'Slices.SErviNGSliCes' since 'SErviNGSliCes' is the strongest extension
// (its strength is -1).
// Example:
// >>> Strongest_Extension("my_class", ["AA", "Be", "CC"])
// "my_class.AA"
function Strongest_Extension(class_name, extensions){
var output = class_name + ".";
var min_strength = 1;
for (var i = 0; i < extensions.length; i++) {
var strength = -1*(extensions[i].replace(/[^A-Z]/g, "").length - extensions[i].replace(/[^a-z]/g, "").length);
if (strength < min_strength) {
min_strength = strength;
output = class_name + "." + extensions[i];
}
} | 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 {
// 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;
}
}
} |
from typing import List
def pairs_sum_to_zero(l: List[int]) -> bool:
"""
pairs_sum_to_zero takes a list of integers as an input.
it returns True if there are two distinct elements in the list that
sum to zero, and False otherwise.
>>> pairs_sum_to_zero([1, 3, 5, 0])
False
>>> pairs_sum_to_zero([1, 3, -2, 1])
False
>>> pairs_sum_to_zero([1, 2, 3, 7])
False
>>> pairs_sum_to_zero([2, 4, -5, 3, 5, 7])
True
>>> pairs_sum_to_zero([1])
False
"""
for x in l:
for y in l:
if x+y == 0 and x != y: | 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 {
// You are given an array array list of numbers.
// You need to return the sum of squared numbers in the given array list,
// round each element in the array list to the upper int(Ceiling) first.
// Examples:
// >>> lst((new ArrayList<Float>(Arrays.asList((float)1.0f, (float)2.0f, (float)3.0f))))
// (14l)
// >>> lst((new ArrayList<Float>(Arrays.asList((float)1.0f, (float)4.0f, (float)9.0f))))
// (98l)
// >>> lst((new ArrayList<Float>(Arrays.asList((float)1.0f, (float)3.0f, (float)5.0f, (float)7.0f))))
// (84l)
// >>> lst((new ArrayList<Float>(Arrays.asList((float)1.4f, (float)4.2f, (float)0.0f))))
// (29l)
// >>> lst((new ArrayList<Float>(Arrays.asList((float)-2.4f, (float)1.0f, (float)1.0f))))
// (6l)
public static long sumSquares(ArrayList<Float> lst) {
if (lst.size() == 0) {
return 0;
}
long sum = 0;
for (int i = 0; i < lst.size(); i++) {
sum += (long)Math.ceil(lst.get(i)) * (long)Math.ceil(lst.get(i));
}
return sum;
} | } | |
from typing import List
def separate_paren_groups(paren_string: str) -> List[str]:
""" 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 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.
>>> separate_paren_groups('( ) (( )) (( )( ))')
['()', '(())', '(()())']
"""
paren_string = paren_string.replace(' ', '')
groups = []
group = ''
open_count = 0
for c in paren_string:
if c == '(':
open_count += 1
elif c == ')':
open_count -= 1
group += c
if open_count == 0:
groups.append(group)
group = '' | return groups | |
//I think we all remember that feeling when the result of some long-awaited
// event is finally known. The feelings and thoughts you have at that moment are
// definitely worth noting down and comparing.
// Your task is to determine if a person correctly guessed the results of a number of matches.
// You are given two arrays of scores and guesses of equal length, where each index shows a match.
// Return an array of the same length denoting how far off each guess was. If they have guessed correctly,
// the value is 0, and if not, the value is the absolute difference between the guess and the score.
// example:
// >>> compare([1, 2, 3, 4, 5, 1], [1, 2, 3, 4, 2, -2])
// [0, 0, 0, 0, 3, 3]
// >>> compare([0, 5, 0, 0, 0, 4], [4, 1, 1, 0, 0, -2])
// [4, 4, 1, 0, 0, 6]
function compare(game, guess){
let result = [];
for(let i = 0; i < game.length; i++){
if(game[i] === guess[i]){
result.push(0); | } else { | result.push(Math.abs(game[i] - guess[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 {
// Write a function that accepts an array array list of strings.
// The array list 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.
// >>> findMax((new ArrayList<String>(Arrays.asList((String)"name", (String)"of", (String)"string"))))
// ("string")
// >>> findMax((new ArrayList<String>(Arrays.asList((String)"name", (String)"enam", (String)"game"))))
// ("enam")
// >>> findMax((new ArrayList<String>(Arrays.asList((String)"aaaaaaa", (String)"bb", (String)"cc"))))
// ("aaaaaaa")
public static String findMax(ArrayList<String> words) {
Map<String, Integer> wordsToUniqueCharacters = new HashMap<String, Integer>();
for (String word : words) {
Set<Character> uniqueCharacters = new HashSet<Character>();
for (char c : word.toCharArray()) {
uniqueCharacters.add(c);
}
wordsToUniqueCharacters.put(word, uniqueCharacters.size());
}
words.sort(new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
int compareResult = wordsToUniqueCharacters.get(o2).compareTo(wordsToUniqueCharacters.get(o1));
if (compareResult == 0) {
return o1.compareTo(o2); | } | return compareResult;
}
});
return words.get(0);
}
} |
//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.
function car_race_collision(n){
let num_collisions = 0;
let left_to_right = Array(n).fill(1);
let right_to_left = Array(n).fill(1);
for (let i = 0; i < left_to_right.length; i++) {
for (let j = 0; j < right_to_left.length; j++) {
if (left_to_right[i] == right_to_left[j]) {
num_collisions++;
} | } | }
return num_collisions;
}
|
//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 an array, where element at index
// i represents the number of stones in the level (i+1).
// Examples:
// >>> make_a_pile(3)
// [3, 5, 7]
function make_a_pile(n){
let arr = [];
for(let i = 0; i < n; i++){
arr.push(n + (i * 2)); | } | return arr;
} |
//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;
} |
//You are given a word. Your task is to find the closest vowel that stands between
// two consonants from the right side of the word (case sensitive).
// Vowels in the beginning and ending doesn't count. Return empty string if you didn't
// find any vowel met the above condition.
// You may assume that the given string contains English letter only.
// Example:
// >>> get_closest_vowel("yogurt")
// "u"
// >>> get_closest_vowel("FULL")
// "U"
// >>> get_closest_vowel("quick")
// ""
// >>> get_closest_vowel("ab")
// ""
function get_closest_vowel(word){
let vowels = ["a", "e", "i", "o", "u"];
let i = word.length - 1;
let result = [];
if (word.length === 1) {
return ""
} else {
while (i >= 0) {
if (vowels.includes(word[i])) {
if (i > 0 && i < word.length - 1) {
if (!vowels.includes(word[i-1]) && !vowels.includes(word[i+1])) {
result.push(word[i]);
}
}
}
i--;
} | if (result.length === 0) { | return ""
} else {
return result[0];
}
}
} |
//This function takes an array l and returns an array 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.
// >>> sort_third([1, 2, 3])
// [1, 2, 3]
// >>> sort_third([5, 6, 3, 4, 8, 9, 2])
// [2, 6, 3, 4, 8, 9, 5]
function sort_third(l){
let x=[];
let y=[];
let k=0;
for(let i=0;i<l.length;i++){
if(i%3==0){
y[k]=l[i];
k=k+1;
}
else{
x[i]=l[i];
}
}
y.sort();
k=0; | for(let i=0;i<l.length;i++){ | if(i%3==0){
x[i]=y[k];
k=k+1;
}
}
return x;
} |
//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
function is_multiply_prime(a){
a = Math.abs(a);
const prime = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];
let i = 0;
let check = [];
while(a != 1){
if(a % prime[i] == 0){
a = a / prime[i];
check.push(prime[i]); | }else{ | i++;
}
}
if(check.length == 3){
return true;
}else{
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 {
// 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;
}
} |
//Given an array of numbers, return the sum of squares of the numbers
// in the array that are odd. Ignore numbers that are negative or not integers.
// >>> double_the_difference([1, 3, 2, 0])
// 10
// >>> double_the_difference([-1, -2, 0])
// 0
// >>> double_the_difference([9, -2])
// 81
// >>> double_the_difference([0])
// 0
// If the input array is empty, return 0.
function double_the_difference(lst){
sum = 0;
for (var i = 0; i < lst.length; i++){ | if(lst[i] % 2 == 1 && lst[i] > 0){ | sum += lst[i] ** 2;
}
}
return sum;
}
|
//You are given an array of integers.
// Write a function next_smallest() that returns the 2nd smallest element of the array.
// Return undefined if there is no such element.
// >>> next_smallest([1, 2, 3, 4, 5])
// 2
// >>> next_smallest([5, 1, 4, 3, 2])
// 2
// >>> next_smallest([])
// undefined
// >>> next_smallest([1, 1])
// undefined
function next_smallest(lst){
if(lst.length<=1){
return undefined;
}
var smallest=Math.min(...lst);
var second_smallest=Number.MAX_VALUE;
for(var i in lst){
if(lst[i]>smallest&&lst[i]<second_smallest){
second_smallest=lst[i];
}
} | if(second_smallest==Number.MAX_VALUE){ | return undefined;
}
return second_smallest;
} |
//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 of GPAs for some students and you have to write
// a function that can output an array 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-"]
function numerical_letter_grade(grades){
var grades_array = [];
for (var i = 0; i < grades.length; i++){
if (grades[i] >= 4.0){
grades_array.push("A+");
}
else if (grades[i] > 3.7){
grades_array.push("A");
}
else if (grades[i] > 3.3){
grades_array.push("A-");
}
else if (grades[i] > 3.0){
grades_array.push("B+");
}
else if (grades[i] > 2.7){
grades_array.push("B");
}
else if (grades[i] > 2.3){
grades_array.push("B-");
}
else if (grades[i] > 2.0){
grades_array.push("C+");
}
else if (grades[i] > 1.7){
grades_array.push("C");
}
else if (grades[i] > 1.3){
grades_array.push("C-");
}
else if (grades[i] > 1.0){
grades_array.push("D+");
}
else if (grades[i] > 0.7){
grades_array.push("D");
}
else if (grades[i] > 0.0){
grades_array.push("D-");
}
else if (grades[i] == 0.0){
grades_array.push("E");
}
}
return grades_array; | } | |
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 non-empty array of integers, return the sum of all of the odd elements that are in even positions.
// Examples
// >>> solution([5, 8, 7, 1])
// 12
// >>> solution([3, 3, 3, 3, 3])
// 9
// >>> solution([30, 13, 24, 321])
// 0
function solution(lst){
let result = 0;
for (let i = 0; i < lst.length; i++){
if (i % 2 === 0){
if (lst[i] % 2 === 1){
result += lst[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 {
// Given an array array list of numbers, return whether or not they are sorted
// in ascending order. If array list has more than 1 duplicate of the same
// number, return false. Assume no negative numbers and only integers.
// Examples
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)5l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)2l, (long)4l, (long)5l))))
// (false)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l, (long)6l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)3l, (long)4l, (long)5l, (long)6l, (long)7l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)3l, (long)2l, (long)4l, (long)5l, (long)6l, (long)7l))))
// (false)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)2l, (long)3l, (long)3l, (long)4l))))
// (true)
// >>> isSorted((new ArrayList<Long>(Arrays.asList((long)1l, (long)2l, (long)2l, (long)2l, (long)3l, (long)4l))))
// (false)
public static boolean isSorted(ArrayList<Long> lst) {
if (lst.size() == 0) {
return true;
}
else {
int running_max = 0;
int previous_number = 0;
boolean first_time = true;
boolean seen_2_in_a_row = false;
for (Long number : lst) {
if (first_time) {
first_time = false;
previous_number = number.intValue();
running_max = number.intValue();
}
else { | if (running_max > number.intValue()) { | return false;
}
else {
if (previous_number == number.intValue()) {
if (seen_2_in_a_row) {
return false;
}
else {
seen_2_in_a_row = true;
}
}
else {
seen_2_in_a_row = false;
}
previous_number = number.intValue();
running_max = number.intValue();
}
}
}
return true;
}
}
} |
//"
// This function will take an array of integers. For all entries in the array, the function shall square the integer entry if its index is a
// multiple of 3 and will cube the integer entry if its index is a multiple of 4 and not a multiple of 3. The function will not
// change the entries in the array whose indexes are not a multiple of 3 or 4. The function shall then return the sum of all entries.
// Examples:
// >>> lst
// [1, 2, 3]
// >>> lst
// []
// >>> lst
// [-1, -5, 2, -1, -5]
function sum_squares(lst){
var sum = 0;
var len = lst.length;
for (var i = 0; i < len; i++){
if (i % 3 == 0){
sum += lst[i]*lst[i];
} else if (i % 4 == 0){
sum += lst[i]*lst[i]*lst[i];
} else {
sum += lst[i];
}
}
return sum; | } | |
//Concatenate array of strings into a single string
// >>> concatenate([])
// ""
// >>> concatenate(["a", "b", "c"])
// "abc"
function concatenate(strings){
if(strings.length === 0){
return "";
}
var string = strings.join("");
return string; | } | |
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;
} | } | |
//You will be given a string of words separated by commas or spaces. Your task is
// to split the string into words and return an array of the words.
// For example:
// >>> words_string("Hi, my name is John")
// ["Hi", "my", "name", "is", "John"]
// >>> words_string("One, two, three, four, five, six")
// ["One", "two", "three", "four", "five", "six"]
function words_string(s){
let res = s.split(/[\s,]+/).filter(Boolean);
return res; | } | |
//The Brazilian factorial is defined as:
// brazilian_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!
// where n > 0
// For example:
// >>> special_factorial(4)
// 288
// The function will receive an integer as input and should return the special
// factorial of this integer.
function special_factorial(n){
if (n > 0) {
let fact = 1;
for (let i = 1; i <= n; i++) { | fact *= i; | }
return fact * special_factorial(n - 1);
} else {
return 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 {
// For a given number n, find the largest number that divides n evenly, smaller than n
// >>> largestDivisor((15l))
// (5l)
public static long largestDivisor(long n) {
for (long i = 2; i <= Math.floor(Math.sqrt(n)); i++) {
if (n % i == 0) {
return n / i; | } | }
return 1;
}
} |
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
|
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;
}
} |
//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;
} |
from typing import List
def will_it_fly(q: List[int], w: int) -> bool:
"""
Write a function that returns True if the object q will fly, and False otherwise.
The object q will fly if it's balanced (it is a palindromic list) and the sum of its elements is less than or equal the maximum possible weight w.
Example:
>>> will_it_fly([1, 2], 5)
False
# 1+2 is less than the maximum possible weight, but it's unbalanced.
>>> will_it_fly([3, 2, 3], 1)
False
# it's balanced, but 3+2+3 is more than the maximum possible weight.
>>> will_it_fly([3, 2, 3], 9)
True
# 3+2+3 is less than the maximum possible weight, and it's balanced.
>>> will_it_fly([3], 5)
True
# 3 is less than the maximum possible weight, and it's balanced.
"""
| s = sum(q) | return s <= w and q == q[::-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 {
// The FibFib number sequence is a sequence similar to the Fibbonacci sequnece that's defined as follows:
// fibfib(0) == 0
// fibfib(1) == 0
// fibfib(2) == 1
// fibfib(n) == fibfib(n-1) + fibfib(n-2) + fibfib(n-3).
// Please write a function to efficiently compute the n-th element of the fibfib number sequence.
// >>> fibfib((1l))
// (0l)
// >>> fibfib((5l))
// (4l)
// >>> fibfib((8l))
// (24l)
public static long fibfib(long n) {
if (n == 0) {
return 0;
}
if (n == 1) {
return 0;
}
if (n == 2) {
return 1;
}
long[] memo = new long[(int) n + 1];
memo[0] = 0;
memo[1] = 0;
memo[2] = 1;
for (int i = 3; i <= n; i++) {
memo[i] = memo[i - 1] + memo[i - 2] + memo[i - 3];
} | return memo[(int) n]; | }
} |
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 the largest prime factor of n. Assume n > 1 and is not a prime.
// >>> largestPrimeFactor((13195l))
// (29l)
// >>> largestPrimeFactor((2048l))
// (2l)
public static long largestPrimeFactor(long n) {
List<Long> primes = new ArrayList<>();
long x = n;
for (long i = 2; i * i <= x; i++) {
if (x % i == 0) {
primes.add(i);
x /= i;
i--;
}
}
if (x > 1) {
primes.add(x);
}
return primes.get(primes.size() - 1); | } | } |
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
|
//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;
}
|
from typing import List
def specialFilter(nums: List[int]) -> int:
"""Write a function that takes an array of numbers as input and returns
the number of elements in the array that are greater than 10 and both
first and last digits of a number are odd (1, 3, 5, 7, 9).
For example:
>>> specialFilter([15, -73, 14, -15])
1
>>> specialFilter([33, -2, -3, 45, 21, 109])
2
"""
count = 0
for num in nums:
if num < 10:
continue
else: | rev = int(str(num)[::-1]) | if num % 2 != 0 and rev % 2 != 0:
count += 1
return count
|
def words_in_sentence(sentence: str) -> str:
"""
You are given a string representing a sentence,
the sentence contains some words separated by a space,
and you have to return a string that contains the words from the original sentence,
whose lengths are prime numbers,
the order of the words in the new string should be the same as the original one.
Example 1:
>>> words_in_sentence('This is a test')
'is'
Example 2:
>>> words_in_sentence('lets go for swimming')
'go for'
Constraints:
* 1 <= len(sentence) <= 100
* sentence contains only letters
"""
| def is_prime(n: int) -> bool: | if n <= 1:
return False
for i in range(2, n):
if n % i == 0:
return False
return True
words = sentence.split()
prime_words = []
for word in words:
if is_prime(len(word)):
prime_words.append(word)
return ' '.join(prime_words) |
//For a given number n, find the largest number that divides n evenly, smaller than n
// >>> largest_divisor(15)
// 5
function largest_divisor(n){
let ld = 0;
for(i=1; i<n; i++){
if(n%i===0){
if(i>ld){
ld = i;
}
}
} | return ld; | } |
//Given the lengths of the three sides of a triangle. Return true if the three
// sides form a right-angled triangle, false otherwise.
// A right-angled triangle is a triangle in which one angle is right angle or
// 90 degree.
// Example:
// >>> right_angle_triangle(3, 4, 5)
// true
// >>> right_angle_triangle(1, 2, 3)
// false
function right_angle_triangle(a, b, c){
if(a**2 + b**2 == c**2 || b**2 + c**2 == a**2 || a**2 + c**2 == b**2) { | return true; | } else {
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 {
// You are given 2 words. You need to return true if the second word or any of its rotations is a substring in the first word
// >>> cycpatternCheck(("abcd"), ("abd"))
// (false)
// >>> cycpatternCheck(("hello"), ("ell"))
// (true)
// >>> cycpatternCheck(("whassup"), ("psus"))
// (false)
// >>> cycpatternCheck(("abab"), ("baa"))
// (true)
// >>> cycpatternCheck(("efef"), ("eeff"))
// (false)
// >>> cycpatternCheck(("himenss"), ("simen"))
// (true)
public static boolean cycpatternCheck(String a, String b) {
int len = b.length();
boolean found = false;
for (int i = 0; i < len; i++) {
b = b.substring(1) + b.charAt(0);
found = a.contains(b);
if (found) break;
}
return found;
} | } | |
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 {
// 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)
public static boolean simplify(String x, String n) {
String[] nums = x.split("/"); | String[] n_nums = n.split("/"); | int x_num = Integer.parseInt(nums[0]);
int n_num = Integer.parseInt(n_nums[0]);
int x_denom = Integer.parseInt(nums[1]);
int n_denom = Integer.parseInt(n_nums[1]);
int res_num = x_num * n_num;
int res_denom = x_denom * n_denom;
return (res_num % res_denom == 0);
}
} |
//In this task, you will be given a string that represents a number of apples and oranges
// that are distributed in a basket of fruit this basket contains
// apples, oranges, and mango fruits. Given the string that represents the total number of
// the oranges and apples and an integer that represent the total number of the fruits
// in the basket return the number of the mango fruits in the basket.
// for examble:
// >>> fruit_distribution("5 apples and 6 oranges", 19)
// 8
// >>> fruit_distribution("0 apples and 1 oranges", 3)
// 2
// >>> fruit_distribution("2 apples and 3 oranges", 100)
// 95
// >>> fruit_distribution("100 apples and 1 oranges", 120)
// 19
function fruit_distribution(s, n){
let split_s = s.split(" ");
let num_of_apples = parseInt(split_s[0]);
let num_of_oranges = parseInt(split_s[3]);
let sum = num_of_apples + num_of_oranges;
return n - sum; | } | |
//You are given a string representing a sentence,
// the sentence contains some words separated by a space,
// and you have to return a string that contains the words from the original sentence,
// whose lengths are prime numbers,
// the order of the words in the new string should be the same as the original one.
// Example 1:
// >>> words_in_sentence("This is a test")
// "is"
// Example 2:
// >>> words_in_sentence("lets go for swimming")
// "go for"
// Constraints:
// * 1 <= len(sentence) <= 100
// * sentence contains only letters
function words_in_sentence(sentence){
return sentence.split(' ').filter(word => {
var number = word.length;
var isPrime = true;
if(number === 1 || number === 0){
return false;
}
for(var i = 2; i < number; i++){
if(number % i === 0){ | isPrime = false; | break;
}
}
return isPrime;
}).join(' ');
} |
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 an array array list of numbers as input and returns
// the number of elements in the array array list that are greater than 10 and both
// first and last digits of a number are odd (1, 3, 5, 7, 9).
// For example:
// >>> specialFilter((new ArrayList<Long>(Arrays.asList((long)15l, (long)-73l, (long)14l, (long)-15l))))
// (1l)
// >>> specialFilter((new ArrayList<Long>(Arrays.asList((long)33l, (long)-2l, (long)-3l, (long)45l, (long)21l, (long)109l))))
// (2l)
public static long specialFilter(ArrayList<Long> nums) {
class helper {
public long getFirstDigit(long n) {
long[] digits = getDigits(n);
return digits[0];
}
public long getLastDigit(long n) {
long[] digits = getDigits(n);
return digits[digits.length - 1];
}
public long[] getDigits(long n) {
ArrayList<Long> digits = new ArrayList<Long>();
while (n != 0) {
digits.add(n % 10);
n /= 10;
}
Collections.reverse(digits);
return digits.stream().mapToLong(i -> i).toArray();
}
} | helper h = new helper(); | return nums.stream().filter(x -> x > 10).filter(x -> {
long firstDigit = h.getFirstDigit(x);
long lastDigit = h.getLastDigit(x);
return firstDigit % 2 != 0 && lastDigit % 2 != 0;
}).count();
}
} |
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