> For the complete documentation index, see [llms.txt](https://gl01.gitbook.io/gfg-editorials/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://gl01.gitbook.io/gfg-editorials/2023/09-2023-sep-11/12-perfect-numbers.md).

# 12. Perfect Numbers

The problem can be found at the following link: [Question Link](https://practice.geeksforgeeks.org/problems/perfect-numbers3207/1)

## My Approach

A straightforward brute-force method that involves identifying all factors that divide the given number. Subsequently, these factors are summed, and the resulting sum is compared to `n` to check whether they both are equal.

* To determine if a number is perfect, I iterate through all numbers from 2 to the square root of the given number.
  * For each divisor `i`, I check
    * if it divides `n`. If it does, I add `i` to our sum.
  * Additionally, if `n` is not equal to `i` (to avoid counting the same factor twice)
    * if it I add `n/i` to the sum.
* After looping through all potential divisors, I compare the sum to `n`. If they are equal, the number is perfect.

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(sqrt(n))`, where `n` is the input number. This is because we iterate up to the square root of `n` to find divisors.
* **Auxiliary Space Complexity**: `O(1)`. We use a constant amount of extra space for the `sum` variable.

## Code (C++)

```cpp
class Solution {
public:
    int isPerfectNumber(long long n) {
        if (n == 1)
            return 0;
        long long sum = 1;
        for (int i = 2; i <= sqrt(n); ++i) {
            if (n % i == 0) {
                sum += i;
                if (n / i != i)
                    sum += n / i;
            }
        }
        return sum == n;
    }
};
```

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