# Solving Leetcode Interviews in Seconds with AI: Palindrome Partitioning II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "132" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s.   Example 1:  Input: s = "aab" Output: 1 Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.  Example 2:  Input: s = "a" Output: 0  Example 3:  Input: s = "ab" Output: 1    Constraints:  1 <= s.length <= 2000 s consists of lowercase English letters only.  

	# Explanation
	Here's an efficient solution to the palindrome partitioning problem, along with an explanation:

*   **Palindrome Identification:** Precompute a table indicating whether each substring of `s` is a palindrome. This avoids redundant palindrome checks.
*   **Dynamic Programming:** Use dynamic programming to determine the minimum cuts needed for each prefix of `s`. The minimum cuts for a prefix ending at index `i` is the minimum of (minimum cuts for prefix ending at `j` + 1) for all `j` such that substring `s[j+1...i]` is a palindrome.
*   **Optimization:** The precomputed palindrome table optimizes the overall process significantly.

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

	
	# Code
	```python
	def minCut(s: str) -> int:
    n = len(s)

    # is_palindrome[i][j] is True if s[i:j+1] is a palindrome
    is_palindrome = [[False] * n for _ in range(n)]

    for i in range(n):
        is_palindrome[i][i] = True

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if length == 2:
                is_palindrome[i][j] = (s[i] == s[j])
            else:
                is_palindrome[i][j] = (s[i] == s[j] and is_palindrome[i+1][j-1])

    # dp[i] is the minimum cuts needed for s[:i+1]
    dp = [0] * n
    for i in range(n):
        if is_palindrome[0][i]:
            dp[i] = 0
        else:
            dp[i] = i  # Maximum possible cuts
            for j in range(i):
                if is_palindrome[j+1][i]:
                    dp[i] = min(dp[i], dp[j] + 1)

    return dp[n-1]
	```