findOutlier kata

by trentHarlem

JavaScript

const findOutlier = (integers) => {
  let odd = [], even = []
  let separate = integers.forEach(n => n % 2 === 0 ? even.push(n) : odd.push(n));
  return even.length == 1 ? even[0] : odd[0];
}

//------converted to arrow 

/* const findOutlier = (integers) => {
  let odd = [], even = []
  let separate = integers.forEach(n => n % 2 === 0 ? even.push(n) : odd.push(n));
  return even.length == 1 ? even[0] : odd[0];
} */

// change last if statemet to return ternary operator 
/* function findOutlier(integers) {
  let odd = [], even = []
  let separate = integers.forEach(n => n % 2 === 0 ? even.push(n) : odd.push(n));
  return even.length == 1 ? even[0] : odd[0];
}
 */

//if (odd.length === 1) return odd[0];else return even[0];
//odd.length===1?odd[0]:even[0]

//-----

/* 
function findOutlier(integers){
  //your code here
  let odd = []
  let even = []
  let separate = integers.forEach(n => {
  if(n % 2 === 0) {
  even.push(n)
  } else odd.push(n)
  });
//console.log('odd',odd)
//console.log('even',even)
if (odd.length===1) {
return odd[0]
}
else return even[0];
}
//-----
 */
//console.log(findOutlier([0, 1, 2]), 1)
console.log(findOutlier([1, 2, 3]), 2)
console.log(findOutlier([2, 6, 8, 10, 3]), 3)
console.log(findOutlier([0, 0, 3, 0, 0]), 3)
console.log(findOutlier([1, 1, 0, 1, 1]), 0)


console.log(findOutlier([2, 4, 0, 100, 4, 11, 2602, 36]), 11)
//Should return: 11 (the only odd number)

console.log(findOutlier([160, 3, 1719, 19, 11, 13, -21]), 160)
//Should return: 160 (the only even number)