Sum of Squares of Odd Numbers in an Array in JavaScript

The task involves iterating through the array, identifying odd numbers, squaring them, and summing up the squares. The different approaches to accomplish this task are listed below.

Iterative Approach

In this approach, we iterate through the array using a loop and for each element, check if it’s odd. If it is, we calculate its square and add it to the previously calculated square sum.

Example: The below code will find the sum of squares of odd numbers in an array using the iterative approach in JavaScript.

Javascript

function sumOfSquaresOfOddNumbers(arr) {
    let sum = 0;
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] % 2 !== 0) {
            sum += arr[i] * arr[i];
        }
    }
    return sum;
}

const numbers1 = [1, 2, 3, 4, 5];
const numbers2 = [2, 4, 6, 8, 10];
const numbers3 = [1, 3, 5, 7, 9];
console.log(sumOfSquaresOfOddNumbers(numbers1));
console.log(sumOfSquaresOfOddNumbers(numbers2));
console.log(sumOfSquaresOfOddNumbers(numbers3));

Output

35
0
165

Using Array Methods

In this approach, we use array methods like filter() to filter out odd numbers and reduce() to calculate the sum of squares.

Example: The below code find the sum of squares of odd numbers in an array using array methods in JavaScript.

Javascript

function sumOfSquaresOfOddNumbers(arr) {
    return arr.filter(num => num % 2 !== 0).
        reduce((acc, curr) => acc + curr * curr, 0);
}

const numbers1 = [1, 2, 3, 4, 5];
const numbers2 = [2, 4, 6, 8, 10];
const numbers3 = [1, 3, 5, 7, 9];
console.log(sumOfSquaresOfOddNumbers(numbers1));
console.log(sumOfSquaresOfOddNumbers(numbers2));
console.log(sumOfSquaresOfOddNumbers(numbers3));

Output

35
0
165