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
Example
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("))(())())"))Input
"))(())())"
Output
6