1278. Palindrome Partitioning III

# Time:  O(k * n^2)
# Space: O(n^2)

class Solution(object):
    def palindromePartition(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        # dp1[i][j]: minimum number of changes to make s[i, j] palindrome
        dp1 = [[0]*len(s) for _ in xrange(len(s))]
        for l in xrange(1, len(s)+1):
            for i in xrange(len(s)-l+1):
                j = i+l-1
                if i == j-1:
                    dp1[i][j] = 0 if s[i] == s[j] else 1
                elif i != j:
                    dp1[i][j] = dp1[i+1][j-1] if s[i] == s[j] else dp1[i+1][j-1]+1

        # dp2[d][i]: minimum number of changes to divide s[0, i] into d palindromes
        dp2 = [[float("inf")]*len(s) for _ in xrange(2)]
        dp2[1] = dp1[0][:]
        for d in xrange(2, k+1):
            dp2[d%2] = [float("inf")]*len(s)
            for i in xrange(d-1, len(s)):  
                for j in xrange(d-2, i):
                    dp2[d%2][i] = min(dp2[d%2][i], dp2[(d-1)%2][j]+dp1[j+1][i])
        return dp2[k%2][len(s)-1]

Solution from kamyu104/LeetCode-Solutions · MIT