Introduction

In this blog post, we will explore how to solve the LeetCode problem "1312" 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 s. In one step you can insert any character at any index of the string. Return the minimum number of steps to make s palindrome. A Palindrome String is one that reads the same backward as well as forward. Example 1: Input: s = "zzazz" Output: 0 Explanation: The string "zzazz" is already palindrome we do not need any insertions. Example 2: Input: s = "mbadm" Output: 2 Explanation: String can be "mbdadbm" or "mdbabdm". Example 3: Input: s = "leetcode" Output: 5 Explanation: Inserting 5 characters the string becomes "leetcodocteel". Constraints: 1 <= s.length <= 500 s consists of lowercase English letters.

Explanation

Here's the approach to solve this problem, followed by the Python code:

• Key Idea: The minimum number of insertions to make a string a palindrome is equivalent to the difference between the length of the string and the length of its longest palindromic subsequence (LPS).
• Longest Palindromic Subsequence (LPS): Find the LPS of the given string s. This can be solved efficiently using dynamic programming.

• Calculate Insertions: Subtract the length of the LPS from the length of the original string to get the minimum number of insertions required.

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

Code

    def minInsertions(s: str) -> int:
    """
    Calculates the minimum number of insertions to make a string a palindrome.

    Args:
        s: The input string.

    Returns:
        The minimum number of insertions needed.
    """

    n = len(s)

    # dp[i][j] stores the length of the longest palindromic subsequence of s[i:j+1]
    dp = [[0] * n for _ in range(n)]

    # Base case: single characters are palindromes of length 1
    for i in range(n):
        dp[i][i] = 1

    # Fill the dp table in bottom-up manner
    for cl in range(2, n + 1):  # cl is the length of the substring
        for i in range(n - cl + 1):
            j = i + cl - 1
            if s[i] == s[j]:
                dp[i][j] = dp[i + 1][j - 1] + 2
            else:
                dp[i][j] = max(dp[i][j - 1], dp[i + 1][j])

    # The length of the longest palindromic subsequence is dp[0][n-1]
    lps_length = dp[0][n - 1]

    # The minimum number of insertions is the difference between the length of the string and the length of the LPS
    return n - lps_length