Introduction

In this blog post, we will explore how to solve the LeetCode problem "1147" 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 text. You should split it to k substrings (subtext1, subtext2, ..., subtextk) such that: subtexti is a non-empty string. The concatenation of all the substrings is equal to text (i.e., subtext1 + subtext2 + ... + subtextk == text). subtexti == subtextk - i + 1 for all valid values of i (i.e., 1 <= i <= k). Return the largest possible value of k. Example 1: Input: text = "ghiabcdefhelloadamhelloabcdefghi" Output: 7 Explanation: We can split the string on "(ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi)". Example 2: Input: text = "merchant" Output: 1 Explanation: We can split the string on "(merchant)". Example 3: Input: text = "antaprezatepzapreanta" Output: 11 Explanation: We can split the string on "(a)(nt)(a)(pre)(za)(tep)(za)(pre)(a)(nt)(a)". Constraints: 1 <= text.length <= 1000 text consists only of lowercase English characters.

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Iterative Approach: Iterate through possible substring lengths, starting from 1. For each length, check if the string can be split into palindromic substrings based on the given condition.
• Palindrome Check: Efficiently verify if the substrings formed by the chosen length are indeed palindromic around the center of the main string.

• Maximize k: Keep track of the largest 'k' (number of substrings) found so far and update it whenever a valid and larger 'k' is encountered.

• Runtime Complexity: O(n^2), where n is the length of the input string. Storage Complexity: O(1).

Code

    def longest_palindrome_decomposition(text: str) -> int:
    """
    Finds the largest possible value of k such that the string can be split into k substrings
    where subtexti == subtextk - i + 1 for all valid values of i.
    """
    n = len(text)
    max_k = 1  # Default value if no palindromic decomposition is found (at least the string itself)

    for length in range(1, n // 2 + 1):  # Iterate through possible substring lengths
        k = 0
        valid = True
        for i in range(n // (2 * length)):
            if text[i * length:(i + 1) * length] == text[n - (i + 1) * length:n - i * length]:
                k += 2
            else:
                valid = False
                break

        if valid:
            remaining = n % (2 * length)
            if remaining > 0:
                k += 1
            max_k = max(max_k, k)

    return max_k