Problem URL: Perfect number

We will find all the divisors of the number and then sum them up. If the sum is equal to the number, then it is a perfect number.

class Solution:
    def checkPerfectNumber(self, num: int) -> bool:
        if num <= 1:
            return False
        divisorSum = 1
        for i in range(2, int(sqrt(num)) + 1):
            if num % i == 0:
                divisorSum += i
                divisorSum += num // i
        return divisorSum == num

Time complexity: O(sqrt(n))
Space complexity: O(1)