LeetCode #132: Palindrome Partitioning II

Problem statement

(Source) 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.

For example, given s = “aab”,
Return 1 since the palindrome partitioning [“aa”,”b”] could be produced using 1 cut.

Analysis

$O(n^2)$ time complexity, $O(n^2)$ space complexity.

Solution

def minCut(s):
    """
    :type s: str
    :rtype: int
    """
    n = len(s)

    # dp to construct "truth" table.
    dp = [[False for i in xrange(n)] for j in xrange(n)]
    for i in xrange(n):
        dp[i][i] = True
    for i in xrange(n - 1):
        if s[i] == s[i + 1]:
            dp[i][i + 1] = True
    for length in xrange(3, n + 1):
        for i in xrange(n - length + 1):
            j = i + length - 1
            if s[i] == s[j] and dp[i + 1][j - 1]:
                dp[i][j] = True

    # dp to count min-cuts.
    cuts = [float('inf')] * n
    cuts[0] = 0
    for j in xrange(1, n):
        if dp[0][j]:
            cuts[j] = 0
        else:
            temp = float('inf')
            for i in xrange(j):
                if dp[i + 1][j] and temp > cuts[i] + 1:
                    temp = cuts[i] + 1
            cuts[j] = temp

    return cuts[-1]