To solve this problem, we need to find the minimum number of cuts required to split a given string into palindromic substrings. This problem can be efficiently solved using dynamic programming (DP) with a precomputation step to identify palindromic substrings.

Approach

1. Precompute Palindromic Substrings: First, we create a 2D array is_pal where is_pal[i][j] is True if the substring s[i..j] (inclusive) is a palindrome. This is done using the following observations:

   • A substring of length 1 is always a palindrome.
   • A substring of length 2 is a palindrome if both characters are the same.
   • For longer substrings, a substring s[i..j] is a palindrome if the first and last characters are the same and the substring s[i+1..j-1] is also a palindrome.

2. Dynamic Programming for Minimum Cuts: We use another DP array dp where dp[i] represents the minimum number of cuts needed for the substring s[0..i-1] (1-based index). The key insight here is:

   • dp[0] = -1 (base case: an empty string requires no cuts).
   • For each i (end of the substring), we check all possible starting points j (from 0 to i-1). If s[j..i-1] is a palindrome, then dp[i] can be updated as min(dp[i], dp[j] + 1).

Solution Code

def minCut(s: str) -> int:
    n = len(s)
    if n == 0:
        return 0

    # Precompute is_pal[i][j] which is True if s[i..j] is a palindrome
    is_pal = [[False] * n for _ in range(n)]
    for i in range(n-1, -1, -1):
        for j in range(i, n):
            if i == j:
                is_pal[i][j] = True
            elif j == i + 1:
                is_pal[i][j] = (s[i] == s[j])
            else:
                is_pal[i][j] = (s[i] == s[j]) and is_pal[i+1][j-1]

    # Compute dp array where dp[i] is min cuts for s[0..i-1]
    dp = [float('inf')] * (n + 1)
    dp[0] = -1  # Base case

    for i in range(1, n + 1):
        for j in range(i):
            if is_pal[j][i-1]:
                dp[i] = min(dp[i], dp[j] + 1)

    return dp[n]

Explanation

• Precomputation: The is_pal array is filled in reverse order (from longer substrings to shorter ones) to leverage previously computed results for sub-substrings.
• DP Array: The dp array is built by checking all possible palindromic substrings ending at each position. For each valid palindrome, we update the minimum cuts needed by considering the cuts up to the start of the palindrome plus one additional cut.

This approach has a time complexity of O(n²) (due to the precomputation and DP steps) and a space complexity of O(n²) (for the is_pal array). This is efficient enough for typical input sizes (e.g., up to 1000 characters).

Example: For the string "aab", the minimum cuts are 1 (split into "aa" and "b"). The code correctly computes this by checking all possible palindromic substrings and finding the optimal cuts.

This solution efficiently handles the problem and provides the correct minimum number of cuts for any given string.

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