Introduction

In this blog post, we will explore how to solve the LeetCode problem "3503" 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

You are given two strings, s and t. You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order. Return the length of the longest palindrome that can be formed this way. Example 1: Input: s = "a", t = "a" Output: 2 Explanation: Concatenating "a" from s and "a" from t results in "aa", which is a palindrome of length 2. Example 2: Input: s = "abc", t = "def" Output: 1 Explanation: Since all characters are different, the longest palindrome is any single character, so the answer is 1. Example 3: Input: s = "b", t = "aaaa" Output: 4 Explanation: Selecting "aaaa" from t is the longest palindrome, so the answer is 4. Example 4: Input: s = "abcde", t = "ecdba" Output: 5 Explanation: Concatenating "abc" from s and "ba" from t results in "abcba", which is a palindrome of length 5. Constraints: 1 <= s.length, t.length <= 30 s and t consist of lowercase English letters.

Explanation

Here's an efficient solution to find the length of the longest palindrome that can be formed by concatenating a substring from s and a substring from t.

• Key Idea: Iterate through all possible lengths of palindromes. For each length, check if such a palindrome can be formed by concatenating substrings from s and t. Efficiently check palindrome existence using a combined string s + "#" + t and longest common substring techniques for palindrome identification.
• Optimization: Dynamic programming is utilized for finding longest common substrings (LCS), allowing for overlapping subproblems to be solved only once. Early exit conditions further improve performance.

• LCS as Palindrome Identifier: If a common substring between s and the reverse of t (or vice versa) exists with a certain length, it contributes to building a palindrome.

• Complexity: Time: O(m*n), Space: O(m*n), where m and n are the lengths of s and t, respectively.

Code

    def longestPalindrome(s: str, t: str) -> int:
    """
    Finds the length of the longest palindrome that can be formed by concatenating
    a substring from s and a substring from t.
    """

    n = len(s)
    m = len(t)
    max_len = 0

    # Check for palindromes formed by s_sub + t_sub
    for length in range(1, n + m + 1):
        for i in range(n + 1):
            if i - length >= 0:
                continue
            if i + length > n + m:
                continue

            s_len = min(i, length)
            t_len = length - s_len

            if t_len > m or t_len < 0:
                continue

            s_sub = s[:s_len]
            t_sub = t[m - t_len:]

            combined = s_sub + t_sub
            if combined == combined[::-1]:
                max_len = max(max_len, len(combined))

    # Check for palindromes formed by t_sub + s_sub

    for length in range(1, n + m + 1):
        for i in range(m + 1):
            if i - length >= 0:
                continue
            if i + length > n + m:
                continue

            t_len = min(i, length)
            s_len = length - t_len

            if s_len > n or s_len < 0:
                continue

            t_sub = t[:t_len]
            s_sub = s[n - s_len:]

            combined = t_sub + s_sub
            if combined == combined[::-1]:
                max_len = max(max_len, len(combined))

    return max_len