Longest Valid Parentheses Leetcode Solution

In this post, we are going to solve the Longest Valid Parentheses Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

Problem

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 ')'.

Now, let’s see the leetcode solution of Longest Valid Parentheses Leetcode Solution.

Longest Valid Parentheses Leetcode Solution in Python

class Solution:
  def longestValidParentheses(self, s: str) -> int:
    s2 = ')' + s
    # dp[i] := Length of longest valid parentheses substring of s2[1..i]
    dp = [0] * len(s2)

    for i in range(1, len(s2)):
      if s2[i] == ')' and s2[i - dp[i - 1] - 1] == '(':
        dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2

    return max(dp)

Longest Valid Parentheses Leetcode Solution in CPP

class Solution {
 public:
  int longestValidParentheses(string s) {
    const string s2 = ")" + s;
    // dp[i] := Length of longest valid parentheses substring of s2[1..i]
    vector<int> dp(s2.length());

    for (int i = 1; i < s2.length(); ++i)
      if (s2[i] == ')' && s2[i - dp[i - 1] - 1] == '(')
        dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2;

    return *max_element(begin(dp), end(dp));
  }
};

Longest Valid Parentheses Leetcode Solution in Java

class Solution {
  public int longestValidParentheses(String s) {
    final String s2 = ")" + s;
    // dp[i] := Length of longest valid parentheses substring of s2[1..i]
    int dp[] = new int[s2.length()];

    for (int i = 1; i < s2.length(); ++i)
      if (s2.charAt(i) == ')' && s2.charAt(i - dp[i - 1] - 1) == '(')
        dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2;

    return Arrays.stream(dp).max().getAsInt();
  }
}

Note: This problem Longest Valid Parentheses is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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