divmgl

function solution(A) {
  var L = A.length; // Length of the array
  var X = ((L + 1) * L) / 2; // Sum from 1..L
  var Y = 0; // Sum of evaluated values
  var I = 0; // Counter
  var F = []; // Found elements
  var V = -1; // Container
    
  while (I < L) { // Start searching the array
    V = A[I]; // Get the value

divmgl

#include <iostream>
#include <array>
#include <map>
#include <random>

using std::cout;
using std::endl;
using std::string;
using std::array;

divmgl

def solution(A):
    table = dict()
    for num in A:
        if not table.get(num):
            table[num] = 1
        else:
            table.pop(num)
    return table.keys()[0]

divmgl

def solution(A, K):
    if len(A) == 0:
        return A
    for _ in range(K):
        A.insert(0, A.pop())
    return A

divmgl

import math

def solution(X, Y, D):
    return int(math.ceil(float(Y - X) / float(D)))

divmgl

def solution(A):
    total = sum(A)
    diff = sum(range(1, len(A) + 2))
    return abs(diff - total)

divmgl

def solution(N):
    max = 0
    bits = 0
    bin = "{0:b}".format(N)
    
    for c in bin:
        if c == "0":
            bits += 1
        if c == "1":
            if max < bits:

divmgl

function solution(A) {
  var F = []; // Found list
  var I = 0, V = 0; // Counter, container
    
  while (I < A.length) { // Iterate through the array
    V = A[I]; // Store the value
    I++; // Increase counter
    if (F[V]) continue; // If the value exists, continue
    F[V] = true; // Store the value
  }

divmgl

function solution(N, A) {
  var M = A.length; // Length of the entry array
  var C = []; // Counters
  var H = 0; // Highest counter
  var PH = 0; // Previously recorded highest counter
  
  for(K = 0; K < N; K++) { // Initialize the array
    C[K] = 0;
  }
        

divmgl

function solution(X, A) {
  if (A.length === 1) { // If the array is one element
    // And if its first item is 1 as well as the value to search for,
    // the frog doesn't need to move
    if (A[0] === 1 && X === 1) return 0; 
    // If not, it's impossible to go anywhere
    else return -1;
  }
    
  var i = -1, // Counter