Introduction

In this blog post, we will explore how to solve the LeetCode problem "507" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Given an integer n, return true if n is a perfect number, otherwise return false. Example 1: Input: num = 28 Output: true Explanation: 28 = 1 + 2 + 4 + 7 + 14 1, 2, 4, 7, and 14 are all divisors of 28. Example 2: Input: num = 7 Output: false Constraints: 1 <= num <= 108

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-level Approach:

  • Iterate from 1 up to the square root of the input number num.
  • If i is a divisor of num, add both i and num // i to the sum of divisors. Handle the case where i is the square root of num to avoid double-counting.
  • Finally, compare the sum of divisors (excluding num itself) with num to determine if it's a perfect number.

• Complexity:

  • Runtime Complexity: O(sqrt(n))
  • Storage Complexity: O(1)

Code

    def is_perfect_number(num: int) -> bool:
    """
    Given an integer n, return true if n is a perfect number, otherwise return false.

    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.

    Example 1:
    Input: num = 28
    Output: true
    Explanation: 28 = 1 + 2 + 4 + 7 + 14
    1, 2, 4, 7, and 14 are all divisors of 28.

    Example 2:
    Input: num = 7
    Output: false

    Constraints:
    1 <= num <= 108
    """
    if num <= 1:
        return False

    sum_of_divisors = 1
    for i in range(2, int(num**0.5) + 1):
        if num % i == 0:
            sum_of_divisors += i
            if i * i != num:
                sum_of_divisors += num // i

    return sum_of_divisors == num