Introduction

In this blog post, we will explore how to solve the LeetCode problem "2472" 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 a string s and a positive integer k. Select a set of non-overlapping substrings from the string s that satisfy the following conditions: The length of each substring is at least k. Each substring is a palindrome. Return the maximum number of substrings in an optimal selection. A substring is a contiguous sequence of characters within a string. Example 1: Input: s = "abaccdbbd", k = 3 Output: 2 Explanation: We can select the substrings underlined in s = "abaccdbbd". Both "aba" and "dbbd" are palindromes and have a length of at least k = 3. It can be shown that we cannot find a selection with more than two valid substrings. Example 2: Input: s = "adbcda", k = 2 Output: 0 Explanation: There is no palindrome substring of length at least 2 in the string. Constraints: 1 <= k <= s.length <= 2000 s consists of lowercase English letters.

Explanation

Here's the solution to the problem:

• Dynamic Programming: Use dynamic programming to store whether a substring s[i:j+1] is a palindrome and its length. This avoids redundant palindrome checks.

• Greedy Selection: Iterate through the string. If a valid palindrome (length >= k) is found, greedily select it and move the starting position past the selected substring.

• Runtime Complexity: O(n^2), where n is the length of the string.

• Storage Complexity: O(n^2)

Code

    def max_palindromes(s: str, k: int) -> int:
    """
    Finds the maximum number of non-overlapping palindromic substrings of length at least k.

    Args:
        s: The input string.
        k: The minimum length of the palindrome substring.

    Returns:
        The maximum number of non-overlapping palindromes.
    """
    n = len(s)
    dp = [[False] * n for _ in range(n)]

    # All single characters are palindromes of length 1, but they might not be >= k
    for i in range(n):
        dp[i][i] = True

    # Check for palindromes of length 2
    for i in range(n - 1):
        if s[i] == s[i + 1]:
            dp[i][i + 1] = True

    # Check for palindromes of length 3 or greater
    for length in range(3, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if s[i] == s[j] and dp[i + 1][j - 1]:
                dp[i][j] = True

    count = 0
    i = 0
    while i < n:
        longest_palindrome_end = -1
        for j in range(n - 1, i - 1, -1):  # Iterate backwards to find the longest palindrome
            if dp[i][j] and (j - i + 1) >= k:
                longest_palindrome_end = j
                break

        if longest_palindrome_end != -1:
            count += 1
            i = longest_palindrome_end + 1  # Move past the selected palindrome
        else:
            i += 1  # Move to the next character

    return count