Findtheoddint.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 value;
  for(let i = 0; i < A.length; i++){
    let count = 0
    for(let j = 0; j < A.length; j++){
      if(A[i] === A[j]){
        count++
      }   
    }
    if(count % 2 === 1){
      value = A[i]
      break;
    }
  }
  return value
}


//different approach

function findOdd(arr) {
  return arr.find((item, index) => arr.filter(el => el == item).length % 2)
}


//different approach 

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);
  }
}