Introduction

In this blog post, we will explore how to solve the LeetCode problem "2443" 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

Given a non-negative integer num, return true if num can be expressed as the sum of any non-negative integer and its reverse, or false otherwise. Example 1: Input: num = 443 Output: true Explanation: 172 + 271 = 443 so we return true. Example 2: Input: num = 63 Output: false Explanation: 63 cannot be expressed as the sum of a non-negative integer and its reverse so we return false. Example 3: Input: num = 181 Output: true Explanation: 140 + 041 = 181 so we return true. Note that when a number is reversed, there may be leading zeros. Constraints: 0 <= num <= 105

Explanation

Here's the breakdown:

• Iterate and Check: Iterate through potential numbers from 0 up to num. For each number, calculate its reverse and check if their sum equals num.
• Reverse Calculation: Implement a helper function to efficiently reverse a given number.

• Early Exit: If a valid pair is found, immediately return True. If the loop completes without finding a pair, return False.

• Time & Space Complexity: O(N * logN) time complexity, where N is the input number num because reversing a number takes O(log N) time, and O(1) space complexity.

Code

    def is_sum_of_number_and_reverse(num: int) -> bool:
    """
    Given a non-negative integer num, return true if num can be expressed as the sum of any non-negative integer and its reverse, or false otherwise.

    For example:
    is_sum_of_number_and_reverse(443) == True
    is_sum_of_number_and_reverse(63) == False
    is_sum_of_number_and_reverse(181) == True
    """

    def reverse_number(n: int) -> int:
        """Reverses a non-negative integer."""
        reversed_n = 0
        while n > 0:
            reversed_n = reversed_n * 10 + n % 10
            n //= 10
        return reversed_n

    for i in range(num + 1):
        reversed_i = reverse_number(i)
        if i + reversed_i == num:
            return True
    return False