Perfect Numbers in JavaScript

A number is a perfect number if is equal to the sum of its proper divisors, that is, the sum of its positive divisors excluding the number itself. In this article, we will see how to check if a number is a perfect number or not in JavaScript.

Examples:

Input: n = 15
Output: false
Explanation: Divisors of 15 are 1, 3 and 5. Sum of 
divisors is 9 which is not equal to 15.
Input: n = 6
Output: true
Explanation: Divisors of 6 are 1, 2 and 3. Sum of 
divisors is 6.

Approach 1: Naive Approach

• Implement a Guard Clause to check if n is a valid positive integer. If not, return false.
• Initialize a variable sum to 0 to keep track of the sum of proper divisors.
• Iterate through all numbers i from 1 to n - 1.
• For each i, check if it is a divisor of n (i.e., n % i === 0). If i is a divisor, add it to the sum.
• After the loop, compare the sum with the original number n.
• If, sum === n, then n is a perfect number. Output that it’s a perfect number.
• If, sum !== n, then n is not a perfect number. Output that it’s not a perfect number.
• Return the boolean result indicating whether n is a perfect number or not.

Example: In this example, we have used above explained approach.

Javascript

function isPerfectNumber(n) {

    // Guard Clause for non-positive
    // or non-integer input
    if (!Number.isInteger(n) || n <= 0) {
        console.log("Please provide a valid positive integer.");
        return false;
    }

    let sum = 0;

    // Loop to find proper divisors
    // and calculate the sum
    for (let i = 1; i < n; i++) {
        if (n % i === 0) {
            sum += i;
        }
    }

    // Check if the sum of proper divisors
    // equals the original number
    const isPerfect = sum === n;

    // Output the result
    if (isPerfect) {
        console.log(`${n} is a perfect number.`);
    } else {
        console.log(`${n} is not a perfect number.`);
    }

    return isPerfect;
}

// Example 1
isPerfectNumber(25);

// Example 2
isPerfectNumber(6);

Output

25 is not a perfect number.
6 is a perfect number.

Time Complexity: O(n)
Auxiliary Space: O(1)

Approach 2: Optimal Approach

This approach is same as above approach but the loop runs till sqrt(n) instead of n - 1, resulting in a more efficient algorithm for finding divisors and improving the overall time complexity.

Example: In this example, we have used above explained approach.

Javascript

function isPerfectNumber(n) {

    // Guard Clause for non-positive
    // or non-integer input
    if (!Number.isInteger(n) || n <= 0) {
        console.log("Please provide a valid positive integer.");
        return false;
    }

    // Start with 1 as all numbers have 1 as a divisor
    let sum = 1;

    // Loop to find proper divisors
    // and calculate the sum
    for (let i = 2; i <= Math.sqrt(n); i++) {
        if (n % i === 0) {
            sum += i;
            if (i !== n / i) {
                sum += n / i;
            }
        }
    }

    // Check if the sum of proper divisors
    // equals the original number
    const isPerfect = sum === n;

    // Output the result
    if (isPerfect) {
        console.log(`${n} is a perfect number.`);
    } else {
        console.log(`${n} is not a perfect number.`);
    }

    return isPerfect;
}

// Example 1
isPerfectNumber(28);

// Example 2
isPerfectNumber(13);

Output

28 is a perfect number.
13 is not a perfect number.

Time Complexity: O(√n)
Auxiliary Space: O(1)