Perfect Numbers

Problem Description

Given a number N, check if a number is perfect or not. A number is said to be perfect if sum of all its factors excluding the number itself is equal to the number. Return 1 if the number is Perfect otherwise return 0.

Examples

Example 1:

Input:
N = 6
Output:
1 
Explanation:
Factors of 6 are 1, 2, 3 and 6.
Excluding 6 their sum is 6 which
is equal to N itself. So, it's a
Perfect Number.

Example 2:

Input:
N = 10
Output:
0
Explanation:
Factors of 10 are 1, 2, 5 and 10.
Excluding 10 their sum is 8 which
is not equal to N itself. So, it's
not a Perfect Number.

Your Task

You don't need to read input or print anything. Your task is to complete the function isPerfectNumber() which takes an Integer N as input and returns 1 if N is a Perfect number else returns 0.

Expected Time Complexity: $O(sqrt(N))$

Expected Auxiliary Space: $O(1)$

Constraints

• 1 ≤ n ≤ 10^12

Problem Explanation

Given a number N, check if a number is perfect or not. A number is said to be perfect if sum of all its factors excluding the number itself is equal to the number. Return 1 if the number is Perfect otherwise return 0.

Code Implementation

Python

Written by @Ishitamukherjee2004

def is_perfect(N):
sum = 0
for i in range(1, N):
  if N % i == 0:
    sum += i
return (sum == N) and 1 or 0

C++

Written by @Ishitamukherjee2004

int isPerfect(int N) {
int sum = 0;
for (int i = 1; i < N; i++) {
  if (N % i == 0) {
    sum += i;
  }
}
return (sum == N) ? 1 : 0;
}

Javascript

Written by @Ishitamukherjee2004

function isPerfect(N) {
let sum = 0;
for (let i = 1; i < N; i++) {
  if (N % i === 0) {
    sum += i;
  }
}
return (sum === N) ? 1 : 0;
}

Typescript

Written by @Ishitamukherjee2004

function isPerfect(N) {
let sum = 0;
for (let i = 1; i < N; i++) {
  if (N % i === 0) {
    sum += i;
  }
}
return (sum === N) ? 1 : 0;
}

Java

Written by @Ishitamukherjee2004

public int isPerfect(int N) {
int sum = 0;
for (int i = 1; i < N; i++) {
  if (N % i == 0) {
    sum += i;
  }
}
return (sum == N) ? 1 : 0;
}

Solution Logic:

The solution iterates through all numbers from 1 to N-1 and checks if the current number is a factor of N. If it is, it adds the number to the sum. Finally, it checks if the sum is equal to N. If it is, the number is perfect, and the function returns 1. Otherwise, it returns 0.

Time Complexity

• The function operates in constant time, $O(N)$ where N is the input number. This is because the solution iterates through all numbers from 1 to N-1.

Space Complexity

• The function uses constant space, $O(1)$, because it only uses a few extra variables to store the sum and the current number.