findOdd.js

// Given an array of integers, find the one that appears an odd number of times.

// There will always be only one integer that appears an odd number of times.

// Examples
// [7] should return 7, because it occurs 1 time (which is odd).
// [0] should return 0, because it occurs 1 time (which is odd).
// [1,1,2] should return 2, because it occurs 1 time (which is odd).
// [0,1,0,1,0] should return 0, because it occurs 3 times (which is odd).
// [1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).


function findOdd(A) {
  let check = {};

  for (let num of A) {
    check[num] = (check[num] || 0) + 1;
  }

  for (let key in check) {
    if (check[key] % 2 !== 0) {
      return Number(key); // Convert string key back to number
    }
  }
}