Question

Given a string containing just the characters ‘(’ and ‘)’, find the length of the longest valid (well-formed) parentheses substring.

For “(()”, the longest valid parentheses substring is “()”, which has length = 2.

Another example is “)()())”, where the longest valid parentheses substring is “()()”, which has length = 4.

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Hide Similar Problems (E) Valid Parentheses

Solution 1

time complexity: O(n), traverse one time
space complexity: at most O(n)

get idea from here.

the great one

class Solution(object):def longestValidParentheses(self, s):""":type s: str:rtype: int"""stack = []maxnum, start = 0, 0          # start records the start ind of this sequencefor ind,elem in enumerate(s):if elem=="(":stack.append(ind)else:# elem==')'if stack!=[]:stack.pop()if stack==[]:maxnum = max( maxnum, ind-start+1)else:maxnum = max( maxnum, ind-stack[-1])else:                         start = ind + 1   # the sequence is broken, start it againreturn maxnum

Solution 2

Using dynamic programming
time complexity: O(n), space complexity: O(n)

This idea is from here. He uses one list in which res[i] records the number of longest sequence with the last elem s[i-1]. However he didn’t notice there is only one valid longest sequence before any elem since we have merged all possible consecutive sequence once. The simpler solution is at here.

Code

class Solution(object):def longestValidParentheses(self, s):""":type s: str:rtype: int"""res = [0]*(len(s)+1)  # res[i] records the longest sequence with the last elem i maxnum = 0for i in range(1,len(s)+1):j = (i-1) - res[i-1] - 1   # find the previous valid available indexif s[i-1]=="(" or j<0 or s[j]==')':res[i] = 0else:# the longest number before j (last elem of res[j] is s[j-1]) + s[j] + dp[i-1] + s[i-1]res[i] = res[j] + 1 + res[i-1] + 1  maxnum = max( maxnum, res[i])return maxnum

For dynamic programming, we can give different meaning to list according to our algorithm. In this problem, I thought of one way to make s[i] be the number of longest sequence before s[i-1], but have no result. The smart way is to make it be the number with the last elem s[i-1]

Solution 3

time complexity: O(n) , traverse two times
space complextiy: O(1),
get idea from here.

class Solution(object):def longestValidParentheses(self, s):""":type s: str:rtype: int"""maxnum ,cur, start = 0, 0, -1for ind,elem in enumerate(s):if elem=='(':cur += 1else:cur -= 1if cur<0:start, cur = ind, 0elif cur==0:maxnum = max( maxnum, ind-start )s = list(s)s = "".join(reversed(s))cur, start = 0, -1for ind,elem in enumerate(s):if elem==')':cur += 1else:cur -= 1if cur<0:start, cur = ind, 0elif cur==0:maxnum = max( maxnum, ind-start)return maxnum

It is easy to get the first part of the code above. However, it can’t work for the case “()(()”. I stunk on it for a long time. A cleverer way is to do the same thing in a reversed order. Why ? ‘(’ and ‘)’ are symmetry.

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