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Perfect Number

Difficulty: Easy

Problem Description

A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. Given an integer n, return true if n is a perfect number, otherwise return false.

Key Insights

  • A perfect number is defined by the equality of the number and the sum of its proper divisors.
  • Proper divisors of a number can be found by iterating up to its square root, which optimizes the process of finding divisors.
  • If n is a perfect number, it can be expressed as the sum of divisors other than itself.

Space and Time Complexity

Time Complexity: O(sqrt(n)) - We only need to iterate up to the square root of n to find its divisors. Space Complexity: O(1) - We use a constant amount of space.

Solution

To determine if a number n is a perfect number, we will:

  1. Initialize a sum variable to hold the sum of divisors.
  2. Iterate from 1 to the square root of n to find divisors.
  3. For each divisor found, add it to the sum and also include its pair (n/divisor), making sure to avoid adding n itself.
  4. Finally, compare the sum of the divisors to n and return true if they are equal, otherwise return false.

Code Solutions

def checkPerfectNumber(num):
    if num <= 1:
        return False
    sum_of_divisors = 1  # Start with 1 as a proper divisor
    for i in range(2, int(num**0.5) + 1):
        if num % i == 0:  # If i is a divisor
            sum_of_divisors += i
            if i != num // i:  # Avoid adding the square root twice
                sum_of_divisors += num // i
    return sum_of_divisors == num
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