Introduction

In this blog post, we will explore how to solve the LeetCode problem "564" 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 string n representing an integer, return the closest integer (not including itself), which is a palindrome. If there is a tie, return the smaller one. The closest is defined as the absolute difference minimized between two integers. Example 1: Input: n = "123" Output: "121" Example 2: Input: n = "1" Output: "0" Explanation: 0 and 2 are the closest palindromes but we return the smallest which is 0. Constraints: 1 <= n.length <= 18 n consists of only digits. n does not have leading zeros. n is representing an integer in the range [1, 1018 - 1].

Explanation

Here's a breakdown of the solution:

• Palindrome Generation: The core idea is to generate potential palindrome candidates that are close to the input number n. This involves considering the prefix of n and manipulating it to create palindromes that are slightly smaller, equal, or larger than the original number.
• Edge Cases: Handle edge cases carefully, such as when the input is a single digit or when generating palindromes results in numbers with different lengths (e.g., decreasing "100" leads to "99").

• Comparison: Compare the generated palindromes with the original number and select the closest one (preferring the smaller one in case of ties).

• Runtime Complexity: O(log n).

• Storage Complexity: O(log n).

Code

    def nearestPalindromic(n: str) -> str:
    """
    Finds the nearest palindromic integer to a given integer represented as a string.

    Args:
        n: The string representation of the integer.

    Returns:
        The string representation of the nearest palindromic integer.
    """

    length = len(n)
    num = int(n)

    # Handle the trivial case
    if num == 0 or num == 1:
        return "0"

    if num <= 10:
        return str(num-1)

    candidates = []

    # Case 1: One less than the smallest number with the same number of digits
    candidates.append(str(10 ** (length - 1) - 1))

    # Case 2: One more than the largest number with the same number of digits
    candidates.append(str(10 ** length + 1))

    # Case 3: Palindromes derived from the prefix
    prefix = n[:(length + 1) // 2]
    p = int(prefix)

    for x in [p - 1, p, p + 1]:
        prefix_str = str(x)
        if length % 2 == 0:
            candidate = prefix_str + prefix_str[::-1]
        else:
            candidate = prefix_str + prefix_str[:-1][::-1]
        candidates.append(candidate)

    # Find the closest palindrome
    closest = None
    min_diff = float('inf')

    for candidate in candidates:
        if candidate == n:
            continue

        diff = abs(int(candidate) - num)

        if diff < min_diff:
            min_diff = diff
            closest = candidate
        elif diff == min_diff:
            closest = min(closest, candidate)

    return closest