Mastering Longest Valid Parentheses in Rust: The Optimized Two-Pass Approach

Parentheses validation problems are a staple of coding interviews, especially when working with strings. Among them, finding the longest valid parentheses substring is one of the most intriguing challenges, requiring a good balance of logic and efficiency. This article dives into solving this problem in Rust using an optimized two-pass approach, with clear explanations and examples.

The Problem Statement

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

Example:

• Input: "(()))())("
• Output: 4 (The longest valid substring is "(()))" or "())".)

The Two-Pass Approach: An Optimized Solution

To solve this problem efficiently, we’ll use a two-pass algorithm. Instead of relying on extra memory like a dynamic programming table or stack, we use two counters (open and close) to track parentheses in a single pass from left to right, and then another pass from right to left.

The Core Idea

1. Count ( as "open" and ) as "close" using two counters.
2. When the open and close counters match, we know we’ve encountered a valid substring.
3. Reset counters when mismatches occur:

• In the left-to-right pass, reset when close > open.
• In the right-to-left pass, reset when open > close.

This ensures that we account for both left-heavy and right-heavy mismatched cases.

Rust Implementation

Here’s the optimized Rust solution:

impl Solution {
    pub fn longest_valid_parentheses(s: String) -> i32 {
        let mut ans = 0;
        let mut open = 0;
        let mut close = 0;
        // First pass: Left to right
        for ch in s.chars() {
            if ch == '(' {
                open += 1;
            } else {
                close += 1;
            }
            if open == close {
                ans = ans.max(2 * close);
            } else if close > open {
                open = 0;
                close = 0;
            }
        }
        // Second pass: Right to left
        open = 0;
        close = 0;
        for ch in s.chars().rev() {
            if ch == '(' {
                open += 1;
            } else {
                close += 1;
            }
            if open == close {
                ans = ans.max(2 * open);
            } else if open > close {
                open = 0;
                close = 0;
            }
        }
        ans
    }
}

Step-by-Step Explanation

Let’s break down how the two-pass approach works.

First Pass: Left to Right

1. Traverse the string from left to right.
2. Count ( with the open counter and ) with the close counter.
3. If open == close, calculate the valid substring length: 2 * close.
4. If close > open, reset both counters to 0 because the substring is invalid beyond this point.

Second Pass: Right to Left

1. Traverse the string from right to left.
2. Again, count ( with the open counter and ) with the close counter.
3. If open == close, calculate the valid substring length: 2 * open.
4. If open > close, reset both counters to 0 to handle unmatched (.

This double traversal ensures all valid substrings are captured, including cases where ) appear before (.

Complexity Analysis

1. Time Complexity: O(n) for both passes, where n is the length of the string. This gives an overall complexity of O(n).
2. Space Complexity: O(1) as we use only two counters (open and close).

Comparison with Other Approaches

Dynamic Programming Approach

The DP approach involves creating an array dp where each element stores the length of the longest valid substring ending at that index.

• Time Complexity: O(n)
• Space Complexity: O(n) (extra DP array)
• Code Complexity: Requires handling edge cases with array indexing.

Stack-Based Approach

The stack approach uses a stack to track unmatched indices of parentheses.

• Time Complexity: O(n)
• Space Complexity: O(n) (stack storage)
• Code Complexity: Intuitive but requires careful handling of indices.

Two-Pass Approach

The two-pass approach simplifies logic by eliminating auxiliary data structures.

• Time Complexity: O(n)
• Space Complexity: O(1) (counters only)
• Code Complexity: Cleaner and easier to implement.

Example Walkthrough

Let’s walk through an example: s = "(()))())(".

First Pass (Left to Right):

• Index 0–1: open = 2, close = 0.
• Index 2–3: open == close → Valid substring: 2 * close = 4.
• Index 4–5: close > open. Reset counters.
• Result after first pass: ans = 4.

Second Pass (Right to Left):

• Index 7–8: open = 0, close = 2.
• Index 6–5: open == close → Valid substring: 2 * open = 2.
• Index 4–3: open > close. Reset counters.
• Result after second pass: ans = 4.

Final result: ans = 4.

Edge Cases

1. Empty String: Input: "" Output: 0.
2. No Valid Substring: Input: "(((((((" Output: 0.
3. All Valid Parentheses: Input: "((()))" Output: 6.
4. Mixed Parentheses: Input: "(()))())(()" Output: 6.

Key Takeaways

1. Optimized for Efficiency: The two-pass approach minimizes space usage while ensuring accuracy.
2. Handles Edge Cases Gracefully: Mismatched or unbalanced parentheses are effectively managed by resetting counters.
3. Simplicity and Clarity: Counters make the code easier to read and debug compared to stacks or dynamic programming.

Engage with Us

We’d love to hear your thoughts on this approach! Have you solved similar problems before? Do you have a favorite technique for handling parentheses-related challenges? Share your experiences and solutions in the comments below. Let’s start a discussion!