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.

describe('Example tests', function() {
  doTest([20,1,-1,2,-2,3,3,5,5,1,2,4,20,4,-1,-2,5], 5);
  doTest([1,1,2,-2,5,2,4,4,-1,-2,5], -1);
  doTest([20,1,1,2,2,3,3,5,5,4,20,4,5], 5);
  doTest([10], 10);
  doTest([1,1,1,1,1,1,10,1,1,1,1], 10);
  doTest([5,4,3,2,1,5,4,3,2,10,10], 1);
});

A)

function findOdd(A) {
  //happy coding!
  // 배열을 하나씩 검사해서 중복되어 나오는 수를 객체에 담은 후 카운트
  // 카운트 된 수가 홀수일 경우 그 수를 리턴
  const obj = {};
  
  for (let i = 0; i<A.length ; i++) {
    let key = A[i]
    if (key in obj) {
      obj[key]++
    }
    else if (!(key in obj)) {
      obj[key] = 1
    }
  }
  
  for (count in obj) {
    if (obj[count]%2 === 1) {
      return +count
    }
  }
  
  
  return 0;
}