Solution to Day 2 of 7 Days of JS

30 November 2019

Let’s see. First of all a solution using a while loop:

function sumboth(array) {
  let count = array.length;
  let i = 0, even = 0, odd = 0;
  while (i < count) {
    array[i] % 2 == 0 ? even += array[i] : odd += array[i];
    i++;  
  }
  return {even, odd};
}

console.log(sumboth([3,1,6,2]))

And as a second choice, a solution using a for loop:

function sumboth2(array) {
  let count = array.length;
  let even = 0, odd = 0;
  for (let i = 0; i < count; i++) {
    array[i] % 2 == 0 ? even += array[i] : odd += array[i];
  }
  return {even, odd};
}

console.log(sumboth2([3,1,4,2]))

I’ve used the conditional (ternary) operator:

condition ? inCaseIsTrue : inCaseIsFalse;

You can read more about it here.

Any questions? Leave a comment!

See you in the next challenge!