Day 2 - Longest valid Parentheses.md

Difficulty	Hard
Source	160 Days of Problem Solving
Tags	Strings Dynamic Programming Stack

🚀 Day 2. Longest valid Parentheses 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a string s consisting of only '(' and ')', find the length of the longest valid parentheses substring.

A parenthesis string is valid if:

1. Every opening parenthesis '(' has a corresponding closing ')'.
2. The closing parenthesis appears after its matching opening parenthesis.

🔍 Example Walkthrough:

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:

2

Explanation:

The longest valid parentheses substring is "()".

Constraints:

• $1 \leq s.length \leq 10^6$
• s consists of '(' and ')' only.

🎯 My Approach:

Stack-Based Approach (O(N) Time, O(N) Space)

1. Use a stack to track indices of parentheses.
2. Push opening parentheses ('(') indices onto the stack.
3. For closing parentheses (')'):
   • If the stack is not empty, pop the top element.
   • If the stack is empty, push the current index as a new base.
   • Maintain the maximum valid length by subtracting indices.

Algorithm Steps:

1. Initialize a stack and push -1 as a base index.
2. Traverse s character by character:
   • If '(', push its index onto the stack.
   • If ')', pop from the stack.
     • If the stack becomes empty, push the current index.
     • Otherwise, update max_length = max(max_length, i - st.top()).
3. Return max_length.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N), as we traverse the string once.
• Expected Auxiliary Space Complexity: O(N), for storing indices in the stack.

📝 Solution Code

Code (C++)

class Solution {
public:
    int maxLength(string s) {
        stack<int> st; st.push(-1);
        int m = 0;
        for (int i = 0; i < s.size(); i++)
            if (s[i] == '(') st.push(i);
            else { 
                st.pop(); 
                if (st.empty()) st.push(i); 
                else m = max(m, i - st.top()); 
            }
        return m;
    }
};

⚡ Alternative Approaches

2️⃣ Two-Pass Counter Approach (O(N) Time, O(1) Space)

1. Use left-right counters to track valid parentheses.
2. Forward pass ensures extra right brackets are ignored.
3. Backward pass ensures extra left brackets are ignored.

class Solution {
public:
    int maxLength(string s) {
        int l = 0, r = 0, m = 0;
        for (char c : s) {
            if (c == '(') l++;
            else r++;
            if (l == r) m = max(m, 2 * r);
            else if (r > l) l = r = 0;
        }
        l = r = 0;
        for (int i = s.size() - 1; i >= 0; i--) {
            if (s[i] == '(') l++;
            else r++;
            if (l == r) m = max(m, 2 * l);
            else if (l > r) l = r = 0;
        }
        return m;
    }
};

🔹 Pros: No extra space needed.
🔹 Cons: Requires two passes.

📊 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Stack (Using Indices)	🟢 O(N)	🟡 O(N)	Simple and effective	Extra stack memory used
Two-Pass Counter Approach	🟢 O(N)	🟢 O(1)	No extra space required	Requires two passes over input

💡 Best Choice?

• ✅ For best efficiency: Two-Pass Counter (O(N)) (No extra space).
• ✅ For simpler implementation: Stack Approach (O(N)) (Easier to understand).

Code (Java)

class Solution {
    static int maxLength(String s) {
        java.util.Stack<Integer> st = new java.util.Stack<>();
        st.push(-1);
        int m = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') st.push(i);
            else {
                st.pop();
                if (st.empty()) st.push(i);
                else m = Math.max(m, i - st.peek());
            }
        }
        return m;
    }
}

Code (Python)

class Solution:
    def maxLength(self, s):
        st=[-1]; m=0
        for i,c in enumerate(s):
            if c=='(':
                st.append(i)
            else:
                st.pop()
                if not st: st.append(i)
                else: m = max(m, i - st[-1])
        return m

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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