Perfect Number

Problem Statement

A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. A positive integer n is given, and we need to determine whether it is a perfect number.

Example 1:

Input: num = 28

Output: true

Explanation: 28 = 1 + 2 + 4 + 7 + 14

Example 2:

Input: num = 6

Output: true

Explanation: 6 = 1 + 2 + 3

Example 3:

Input: num = 496

Output: true

Example 4:

Input: num = 8128

Output: true

Example 5:

Input: num = 2

Output: false

Constraints:

• 1 <= num <= 10^8

Solutions

Approach

To determine if a number is perfect, we can follow these steps:

1. Initialize Sum:

   • Initialize a variable to store the sum of divisors.

2. Find Divisors:

   • Iterate from 1 to num/2 and add all divisors of num to the sum.

3. Check Sum:

   • Compare the sum of divisors with the original number. If they are equal, the number is perfect.

Java

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

C++

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;
    }
};

Python

class Solution:
    def checkPerfectNumber(self, num: int) -> bool:
        if num <= 1:
            return False
        sum_divisors = 1
        for i in range(2, int(num ** 0.5) + 1):
            if num % i == 0:
                sum_divisors += i
                if i != num // i:
                    sum_divisors += num // i
        return sum_divisors == num

These solutions efficiently determine if a number is perfect by iterating through potential divisors up to the square root of the number, ensuring an optimal time complexity.