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Suppose we have a string s. This s consists of opening and closing parenthesis only. We have to find the length of the longest valid (well-formed) parentheses substring. So if the input is like “))(())())”, then the result will be 6, as the valid string is “(())()”.

To solve this, we will follow these steps −

Make a stack, and insert -1., set ans := 0

for i in range 0 to length of stack – 1

if s[i] is opening parentheses, then insert i into stack

otherwise

if stack is not empty and top of stack is not -1 and s[stack top] is opening parentheses, then

top element from stack

ans := max of ans and i – stack top

otherwise insert i into stack

return ans

Let us see the following implementation to get a better understanding −

class Solution(object): def solve(self, s): stack = [-1] ans = 0 for i in range(len(s)): if s[i] == "(": stack.append(i) else: if stack and stack[-1]!=-1 and s[stack[-1]] == "(": stack.pop() ans = max(ans,i - stack[-1]) else: stack.append(i) return ans ob = Solution() print(ob.solve("))(())())"))

"))(())())"

6

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