Q.

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).

A)

function findOdd(A) {
  let newObj = A.reduce((obj, el) => {
    if (el in obj) obj[el] += 1;
    else obj[el] = 1;
    return obj;
  }, {})
  
  for (let el in newObj) {
    if (newObj[el] % 2 === 1)
      return Number(el);
  }
  return 0;
}

other

같은 풀이인데도 이렇게 짧게 줄여쓸 수도 있다.

function findOdd(A) {
  var obj = {};
  A.forEach(function(el){
    obj[el] ? obj[el]++ : obj[el] = 1;
  });
  
  for(prop in obj) {
    if(obj[prop] % 2 !== 0) return Number(prop);
  }
}