1745. Palindrome Partitioning IV

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

class Solution(object):
    def checkPartitioning(self, s):
        """
        :type s: str
        :rtype: bool
        """
        def manacher(s):
            s = '^#' + '#'.join(s) + '#$'
            P = [0]*len(s)
            C, R = 0, 0
            for i in xrange(1, len(s)-1):
                i_mirror = 2*C-i
                if R > i:
                    P[i] = min(R-i, P[i_mirror])
                while s[i+1+P[i]] == s[i-1-P[i]]:
                    P[i] += 1
                if i+P[i] > R:
                    C, R = i, i+P[i]
            return P

        P = manacher(s)
        prefix, suffix = [], []
        for i in xrange(2, len(P)-2):
            if i-1-P[i] == 0:
                prefix.append(i)
            if i+1+P[i] == len(P)-1:
                suffix.append(i)
        for i in prefix:
            for j in suffix:
                left, right = i+1+P[i], j-1-P[j]
                if left > right:
                    continue
                mid = left + (right-left)//2
                if P[mid] >= mid-left:
                    return True
        return False


# Time:  O(n^2)
# Space: O(n^2)
class Solution2(object):
    def checkPartitioning(self, s):
        """
        :type s: str
        :rtype: bool
        """        
        dp = [[False]*len(s) for _ in xrange(len(s))]
        for i in reversed(xrange(len(s))):
            for j in xrange(i, len(s)):
                if s[i] == s[j] and (j-i < 2 or dp[i+1][j-1]):
                    dp[i][j] = True
        for i in xrange(1, len(s)-1):
            if not dp[0][i-1]:
                continue
            for j in xrange(i+1, len(s)):
                if not dp[j][-1]:
                    continue
                if dp[i][j-1]:
                    return True
        return False

Solution from kamyu104/LeetCode-Solutions · MIT