LeetCode-in-All

32. Longest Valid Parentheses

Hard

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

Example 1:

Input: s = “(()”

Output: 2

Explanation: The longest valid parentheses substring is “()”.

Example 2:

Input: s = “)()())”

Output: 4

Explanation: The longest valid parentheses substring is “()()”.

Example 3:

Input: s = “”

Output: 0

Constraints:

• 0 <= s.length <= 3 * 104
• s[i] is '(', or ')'.

To solve the “Longest Valid Parentheses” problem in Swift with a Solution class, we can follow these steps:

1. Define a Solution class.
2. Define a method named longestValidParentheses that takes a string s as input and returns an integer representing the length of the longest valid parentheses substring.
3. Initialize a stack to store the indices of characters.
4. Initialize a variable maxLen to store the maximum length of valid parentheses found so far.
5. Push -1 onto the stack to mark the starting point of a potential valid substring.
6. Iterate through each character of the string:
   • If the character is '(', push its index onto the stack.
   • If the character is ')':
     • Pop the top index from the stack.
     • If the stack is empty after popping, push the current index onto the stack to mark the starting point of the next potential valid substring.
     • Otherwise, update maxLen with the maximum of the current maxLen and i - stack.peek(), where i is the current index and stack.peek() is the index at the top of the stack.
7. Return maxLen.

Here’s the implementation:

class Solution {
    public func longestValidParentheses(_ s: String) -> Int {
        var maxLen = 0
        var left = 0
        var right = 0
        let n = s.count
        let characters = Array(s)

        for i in 0..<n {
            if characters[i] == "(" {
                left += 1
            } else {
                right += 1
            }
            if right > left {
                left = 0
                right = 0
            }
            if left == right {
                maxLen = max(maxLen, left + right)
            }
        }

        left = 0
        right = 0
        for i in (0..<n).reversed() {
            if characters[i] == "(" {
                left += 1
            } else {
                right += 1
            }
            if left > right {
                left = 0
                right = 0
            }
            if left == right {
                maxLen = max(maxLen, left + right)
            }
        }

        return maxLen
    }
}

This implementation provides a solution to the “Longest Valid Parentheses” problem in Swift. It finds the length of the longest valid parentheses substring in the given string s.

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