Approach: Sum divisors up to sqrt(n)

def checkPerfectNumber(n):
    if n <= 1:
        return False
    total, i = 1, 2
    while i * i <= n:
        if n % i == 0:
            total += i
            if i != n // i:
                total += n // i
        i += 1
    return total == n

Complexity: O(sqrt n) time, O(1) space.

Full Solution — all languages

class Solution:
    def checkPerfectNumber(self, n):
        if n <= 1:
            return False
        total, i = 1, 2
        while i * i <= n:
            if n % i == 0:
                total += i
                if i != n // i:
                    total += n // i
            i += 1
        return total == n

import java.util.*;
class Solution {
    public boolean checkPerfectNumber(int n) {
        if (n <= 1) return false;
        int total = 1;
        for (int i = 2; (long) i * i <= n; i++) {
            if (n % i == 0) {
                total += i;
                if (i != n / i) total += n / i;
            }
        }
        return total == n;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: bool checkPerfectNumber(int n) {
        if (n <= 1) return false;
        int total = 1;
        for (int i = 2; (long long) i * i <= n; i++) {
            if (n % i == 0) {
                total += i;
                if (i != n / i) total += n / i;
            }
        }
        return total == n;
    }
};

/**
 * @param {number} n
 * @return {boolean}
 */ var checkPerfectNumber = function(n) {
    if (n <= 1) return false;
    let total = 1;
    for (let i = 2; i * i <= n; i++) {
        if (n % i === 0) {
            total += i;
            if (i !== n / i) total += n / i;
        }
    }
    return total === n;
};

Perfect Number

Example 1

Example 2

Constraints