Perfect Number

Problem

A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. A divisor of an integer x is an integer that can divide x evenly.

Given an integer n, return true if n is a perfect number, otherwise return false.

Example 1: Input: num = 28 Output: true Explanation: 28 = 1 + 2 + 4 + 7 + 14 1, 2, 4, 7, and 14 are all divisors of 28.

Example 2: Input: num = 7 Output: false

Constraints:

• 1 <= num <= 108

Solution

Brute Force - O(n) time - O(1) space

class Solution {
public:
    bool checkPerfectNumber(int num) {      
        int sum = 0;
        for (int i = 1; i < num; i++) {
            if (num % i == 0) sum += i;
        }
        return sum == num;
    }
};

Square Root - O(n) time - O(1) space

Minimize the steps by adding both divisors at once. Minimize search space by going up to the square root of the number.

class Solution {
public:
    bool checkPerfectNumber(int num) {      
        if (num <= 1) return false;
        int sum = 1;
        
        for (int i = 2; i <= sqrt(num); i++) {
            if (num % i == 0) {
                sum += i;
                if (i != num / i) sum += num / i;
            }
        }
        return sum == num;
    }
};