LC-0132 Hard LeetCode

132. Palindrome Partitioning II

Read the full problem statement on LeetCode.

Difficulty: hard Acceptance: 35% Topics: String, Dynamic Programming

Reading material

Strings 12 min 1-D Dynamic Programming 16 min

Reference solution (spoiler · python)

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

class Solution(object):
    # @param s, a string
    # @return an integer
    def minCut(self, s):
        lookup = [[False for j in xrange(len(s))] for i in xrange(len(s))]
        mincut = [len(s) - 1 - i for i in xrange(len(s) + 1)]

        for i in reversed(xrange(len(s))):
            for j in xrange(i, len(s)):
                if s[i] == s[j]  and (j - i < 2 or lookup[i + 1][j - 1]):
                    lookup[i][j] = True
                    mincut[i] = min(mincut[i], mincut[j + 1] + 1)

        return mincut[0]

Solution from kamyu104/LeetCode-Solutions · MIT