LC-0005 Medium LeetCode

5. Longest Palindromic Substring

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 36% Topics: Two Pointers, String, Dynamic Programming

Reading material

Two Pointers 9 min Strings 12 min 1-D Dynamic Programming 16 min

Reference solution (spoiler · python)

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

class Solution(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        def preProcess(s):
            if not s:
                return ['^', '$']
            T = ['^']
            for c in s:
                T +=  ['#', c]
            T += ['#', '$']
            return T

        T = preProcess(s)
        P = [0] * len(T)
        center, right = 0, 0
        for i in xrange(1, len(T) - 1):
            i_mirror = 2 * center - i
            if right > i:
                P[i] = min(right - i, P[i_mirror])
            else:
                P[i] = 0

            while T[i + 1 + P[i]] == T[i - 1 - P[i]]:
                P[i] += 1

            if i + P[i] > right:
                center, right = i, i + P[i]

        max_i = 0
        for i in xrange(1, len(T) - 1):
            if P[i] > P[max_i]:
                max_i = i
        start = (max_i - 1 - P[max_i]) // 2
        return s[start : start + P[max_i]]


# Time:  O(n^2)
# Space: O(1)
class Solution2(object):
    def longestPalindrome(self, s):
        """
        :type s: str
        :rtype: str
        """
        def expand(s, left, right):
            while left >= 0 and right < len(s) and s[left] == s[right]:
                left -= 1
                right += 1
            return (right-left+1)-2
        
        left, right = -1, -2
        for i in xrange(len(s)):
            l = max(expand(s, i, i), expand(s, i, i+1))
            if l > right-left+1:
                right = i+l//2
                left = right-l+1
        return s[left:right+1] if left >= 0 else ""

Solution from kamyu104/LeetCode-Solutions · MIT