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The main idea is that divisors come in pairs, and for each pair (d, n/d), if d is less than the square root of n, then d and n/d are both divisors of n. This approach allows us to only iterate up to the square root of n, making it more efficient.
Time Complexity: O(sqrt(n))
Space Complexity: O(1)
1using System;
2
3public class Solution {
4 public bool CheckPerfectNumber(int num) {
5 if (num <= 1) return false;
6 int sum = 1;
7 int sqrtNum = (int)Math.Sqrt(num);
8 for (int i = 2; i <= sqrtNum; i++) {
9 if (num % i == 0) {
10 sum += i;
11 if (i != num / i) sum += num / i;
12 }
13 }
14 return sum == num;
15 }
16}In C#, we employ the Math.Sqrt function. The same loop logic checks and adds divisors to sum and verifies if sum equates to the input number.
This approach calculates the sum of all divisors excluding the number itself by checking each integer from 1 to n-1 to determine if it's a divisor, then sums these and checks if the sum equals n. This method is simple but inefficient for large numbers.
Time Complexity: O(n)
Space Complexity: O(1)
1public classIn Java, we repeat the naive process. The loop checks each number up to num - 1 for divisibility, adding qualifying divisors to accumulate and finally compare with the original number.