Finding Number That Appears for Odd Times in JavaScript

Given an array of integers, we are required to write a function that takes this array and finds the one element that appears an odd number of times. There will always be only one integer that appears an odd number of times.

We will approach this problem by sorting the array. Once sorted, we can iterate over the array to pick the element that appears for odd number of times.

Example

Following is the code −

const arr = [20, 1, -1, 2, -2, 3, 3, 5, 5, 1, 2, 4, 20, 4, -1, -2, 5];
const findOdd = arr => {
   let count = 0;
   let last;
   arr.sort((a, b) => a - b);
   for (let i = 0; i < arr.length; i++){
      if (arr[i] === last) {
         count++;
         continue;
      };
      if(count % 2){
         return last;
      };
      last = arr[i];
      count = 1;
   };
   return last;
};
console.log(findOdd(arr));

Output

This will produce the following output on console −

5