【LeetCode OJ】Palindrome Partitioning II

Problem Link:

http://oj.leetcode.com/problems/palindrome-partitioning-ii/

We solve this problem by using Dynamic Programming.

Optimal Sub-structure

Assume a string S has the palindrome minimum cuts n, and S = W1 + W2 + ... + Wn where Wi is a palindrome. Then for S' = W1 + W2 + ... + Wn-1, S' must have the palindrome minimum cut n-1. It is easy to prove by the contradiction.

Recursive Formula

Given a string s, let A[0..n-1] be an array where A[i] is the palindrome minimum cuts for s[0..i]. The recursive formula for A[] is:

A[0] = 0, since an empty string is a palindrome

For i > 0, we have

　　A[i] = 0, if s[0..i] is a palindrome

　　A[i] = min{ A[j]+1 | j = 1, ..., i and s[j+1..i] is a palindrome }, otherwise

Implementation

The following code is the python impelmentation accepted by oj.leetcode.com

class Solution:
    # @param s, a string
    # @return an integer
    def minCut(self, s):
        """
        Let A[0..n-1] be a new array, where A[i] is the min-cuts of s[0..i]
        A[0] = 0, since "" is a palindrome
        For i > 0, we have
         A[i] = 0, if s[0..i] is palindrome
         A[i] = min{ A[j]+1 | 0 < j <= i }, otherwise
        """
        n = len(s)
        # n = 0 or 1, return 0, no cut needed
        if n < 2:
            return 0

        # Initialization: s[0..i] at least has i cuts to be partitioned into i characters
        A = range(n)
        for i in xrange(n):
            A[i] = i

        # Compute P: P[i][j] = True if s[i..j] is a palindrome
        P = [None] * n
        for i in xrange(n):
            P[i] = [False] * n

        for mid in xrange(n):
            P[mid][mid] = True
            # Check strings with mid "s[mid]"
            i = mid - 1
            j = mid + 1
            while i >= 0 and j <= n-1 and s[i]==s[j]:
                P[i][j] = True
                i -= 1
                j += 1
            # Check strings with mid "s[mid]s[mid+1]"
            i = mid
            j = mid + 1
            while i >= 0 and j <= n-1 and s[i] == s[j]:
                P[i][j] = True
                i -= 1
                j += 1

        # Quick return, if s[0..n-1] is a palindrome
        if P[0][n-1]:
            return 0

        # DP method, update A from i = 1 to n-1
        for i in xrange(n):
            if P[0][i]:
                A[i] = 0
            else:
                for j in xrange(i):
                    if P[j+1][i]:  # s[0..i] = s[0..j] + s[j+1..i], where s[j+1..i] is a palindrome
                        A[i] = min(A[i], A[j]+1)

        return A[n-1]