Longest Valid Parentheses

LeetCode 32 | Difficulty: Hard

Hard

Problem Description

Given a string containing just the characters '(' and ')', return 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 * 10^4`

- `s[i]` is `'('`, or `')'`.

Topics: String, Dynamic Programming, Stack

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Approach

Dynamic Programming

Break the problem into overlapping subproblems. Define a state (what information do you need?), a recurrence (how does state[i] depend on smaller states?), and a base case. Consider both top-down (memoization) and bottom-up (tabulation) approaches.

When to use

Optimal substructure + overlapping subproblems (counting ways, min/max cost, feasibility).

Stack

Use a stack (LIFO) to track elements that need future processing. Process elements when a "trigger" condition is met (e.g., finding a smaller/larger element). Monotonic stack maintains elements in sorted order for next greater/smaller element problems.

When to use

Matching brackets, next greater element, evaluating expressions, backtracking history.

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

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Solutions

Solution 1: C# (Best: 72 ms)

Metric	Value
Runtime	72 ms
Memory	23.2 MB
Date	2020-01-12

Solution

public class Solution {
    public int LongestValidParentheses(string s) {
            Stack<int> stack = new Stack<int>();
    int result = 0;
    for (int i = 0; i < s.Length; i++)
    {
        if(s[i] == '(')
        {
            stack.Push(i);
        }
        else if(stack.Count >0 && s[stack.Peek()] == '(')
        {
            stack.Pop();
            int validLen = stack.Count == 0 ? i+1 : i- stack.Peek();
            result = Math.Max(result, validLen);
        }
        else
        {
            stack.Push(i);
        }
    }
    
    return result;
    }
}

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Complexity Analysis

Approach	Time	Space
Dynamic Programming	$O(n)$	$O(n)$
Stack	$O(n)$	$O(n)$

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Interview Tips

Key Points

• Break the problem into smaller subproblems. Communicate your approach before coding.
• Define the DP state clearly. Ask: "What is the minimum information I need to make a decision at each step?"
• Consider if you can reduce space by only keeping the last row/few values.
• Think about what triggers a pop: is it finding a match, or finding a smaller/larger element?