Return the sum of all even integers in array squared (javascript)

Question

As the title says, I am trying to take an array of numbers, and then return the sum of all the even numbers in the array squared.

I can get my function to return the sum of all the even numbers in a given array, but cannot get it to square the sum of the even numbers.

Here is what I have so far:

let numStr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

const squareEvenNumbers = (numStr) => {
  let sum = 0;
  for (let i = 0; i < numStr.length; i++) {
    if (numStr[i] % 2 === 0) {
      sum = Math.pow(sum + numStr[i], 2);
    }
  }
  return sum
}

console.log(squareEvenNumbers(numStr));

try sum += Math.pow(numStr[i], 2) :D

Swetank Poddar
– Swetank Poddar

2019-12-29 23:25:09 +00:00
Commented Dec 29, 2019 at 23:25 — Swetank Poddar
– Swetank Poddar, Commented Dec 29, 2019 at 23:25

CertainPerformance · Accepted Answer · 2019-12-30 00:21:27Z

2

You need to raise only the current item in the array to the power of 2, not the whole sum:

let numStr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

const squareEvenNumbers = (numStr) => {
  let sum = 0;
  for (let i = 0; i < numStr.length; i++) {
    if (numStr[i] % 2 === 0) {
    sum += Math.pow(numStr[i], 2);
    }
  }
  return sum
}
 
console.log(squareEvenNumbers(numStr));

Or, more concisely:

let numStr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

const squareEvenNumbers = arr => arr
  .filter(num => num % 2 === 0)
  .reduce((a, num) => a + num ** 2, 0);
 
console.log(squareEvenNumbers(numStr));

Or, to only iterate over the array once:

let numStr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

const squareEvenNumbers = arr => arr
  .reduce((a, num) => a + (num % 2 === 0 && num ** 2), 0);
 
console.log(squareEvenNumbers(numStr));

edited Dec 30, 2019 at 0:21

answered Dec 29, 2019 at 23:23

CertainPerformance

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