Perfect Number

Perfect Number - Problem

Math Easy

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 (with no remainder).

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

Input & Output

Example 1 — First Perfect Number

$ Input: n = 6

› Output: true

💡 Note: The divisors of 6 are 1, 2, and 3. Sum: 1 + 2 + 3 = 6, which equals the number itself.

Example 2 — Second Perfect Number

$ Input: n = 28

› Output: true

💡 Note: The divisors of 28 are 1, 2, 4, 7, and 14. Sum: 1 + 2 + 4 + 7 + 14 = 28.

Example 3 — Not Perfect

$ Input: n = 12

› Output: false

💡 Note: The divisors of 12 are 1, 2, 3, 4, and 6. Sum: 1 + 2 + 3 + 4 + 6 = 16, which is not equal to 12.

Constraints

1 ≤ n ≤ 10⁸

Visualization

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Asked in

G Google 15 M Microsoft 12 A Amazon 8

The key insight is that divisors come in pairs - if i divides n, then n/i also divides n. We only need to check up to √n and add both divisors in each pair. Best approach is the square root optimization with Time: O(√n), Space: O(1).

Common Approaches

✓ Brute Force

⏱️ Time: O(n) Space: O(1)

Iterate through all numbers from 1 to n-1, check if each divides n evenly, and sum all divisors. Compare the sum with n.

Square Root Optimization

⏱️ Time: O(√n) Space: O(1)

Since divisors come in pairs (if i divides n, then n/i also divides n), we only need to check up to √n and add both divisors in each pair.

Brute Force — Algorithm Steps

Iterate from 1 to n-1
Check if each number divides n evenly
Sum all divisors
Return true if sum equals n

Visualization

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Step-by-Step Walkthrough

Check All

Test every number from 1 to n-1

Sum Divisors

Add up all numbers that divide evenly

Compare

Check if sum equals original number

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool solution(int n) {
    if (n <= 1) {
        return false;
    }
    
    int divisorSum = 0;
    for (int i = 1; i < n; i++) {
        if (n % i == 0) {
            divisorSum += i;
        }
    }
    
    return divisorSum == n;
}

int main() {
    int n;
    scanf("%d", &n);
    
    bool result = solution(n);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop from 1 to n-1 checking each number

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track sum

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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