[Dynamic Programming] Maximum Length of Valid Parentheses Substring - Medium

Finding the maximum length of a valid parentheses substring is a well-known challenge in computer science. A valid parentheses substring means a string where every opening parenthesis has a matching closing parenthesis in the right order. We can solve this problem using dynamic programming or stack methods to find the length of the longest valid substring quickly.

In this article, we will look at different ways to solve the Maximum Length of Valid Parentheses Substring problem. We will explain the dynamic programming approach in Java, Python, and C++. We will also discuss stack methods for these languages. Plus, we will show how to use arrays for dynamic programming in Java and Python. At the end, we will answer common questions about this problem and its solutions.

• [Dynamic Programming] Maximum Length of Valid Parentheses Substring Solution Overview
• Dynamic Programming Approach in Java for Maximum Length of Valid Parentheses
• Dynamic Programming Approach in Python for Maximum Length of Valid Parentheses
• Dynamic Programming Approach in C++ for Maximum Length of Valid Parentheses
• Stack Based Approach for Maximum Length of Valid Parentheses in Java
• Stack Based Approach for Maximum Length of Valid Parentheses in Python
• Stack Based Approach for Maximum Length of Valid Parentheses in C++
• Using Array for Dynamic Programming in Java
• Using Array for Dynamic Programming in Python
• Frequently Asked Questions

If we want to learn more about dynamic programming, we can check related topics like the Dynamic Programming Fibonacci Number or the Dynamic Programming Climbing Stairs problem.

Dynamic Programming Approach in Java for Maximum Length of Valid Parentheses

We can solve the problem of finding the maximum length of valid parentheses using dynamic programming in Java. We will create a DP array. This array will help us keep track of the lengths of valid parentheses substrings. The main idea is to go through the string. We will update the DP array based on the closing parentheses we find.

Here is a simple implementation of the dynamic programming approach in Java:

public class MaxValidParentheses {
    public int longestValidParentheses(String s) {
        int n = s.length();
        int[] dp = new int[n];
        int maxLen = 0;

        for (int i = 1; i < n; i++) {
            if (s.charAt(i) == ')') {
                if (s.charAt(i - 1) == '(') {
                    dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
                } else if (i - dp[i - 1] > 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
                    dp[i] = dp[i - 1] + (i >= 2 + dp[i - 1] ? dp[i - 2 - dp[i - 1]] : 0) + 2;
                }
                maxLen = Math.max(maxLen, dp[i]);
            }
        }
        return maxLen;
    }

    public static void main(String[] args) {
        MaxValidParentheses mvp = new MaxValidParentheses();
        String s = "(()())";
        System.out.println("Maximum length of valid parentheses: " + mvp.longestValidParentheses(s));
    }
}

Explanation of the Code:

• We create a dp array. The dp[i] stores the length of the longest valid parentheses ending at index i.
• We start going through the string from index 1.
• When we find a closing parenthesis ), we check the character before it:
  • If it is (, we have a valid pair. So we update dp[i].
  • If it is ), we check if the character before the last valid substring is (. Then we update the value accordingly.
• At the end, we keep track of the maximum length we found.

This Java code shows how we can use dynamic programming to find the maximum length of valid parentheses substring. For more reading on dynamic programming, we can check these articles: Dynamic Programming: Fibonacci Number or Dynamic Programming: Longest Valid Parentheses.

Dynamic Programming Approach in Python for Maximum Length of Valid Parentheses

We want to find the maximum length of valid parentheses substring using dynamic programming in Python. We will use a DP array. Each entry in this array shows the length of the longest valid substring that ends at that index.

Algorithm

1. First, we make a DP array. This array will have the same length as the input string and will start with zeros.
2. Next, we loop through the string. We start from the second character.
3. For each character:
   • If it is a closing parenthesis ')', we check the character before it:
     • If the character before is an opening parenthesis '(', we set dp[i] = dp[i-2] + 2.
     • If the character before is also a closing parenthesis ')', we need to find a matching opening parenthesis:
       • If s[i - dp[i-1] - 1] is '(', we set dp[i] = dp[i-1] + dp[i - dp[i-1] - 2] + 2.
4. In the end, we take the maximum value from the DP array.

Python Code

def longestValidParentheses(s: str) -> int:
    n = len(s)
    dp = [0] * n
    max_length = 0

    for i in range(1, n):
        if s[i] == ')':
            if s[i - 1] == '(':
                dp[i] = (dp[i - 2] if i >= 2 else 0) + 2
            elif i - dp[i - 1] - 1 >= 0 and s[i - dp[i - 1] - 1] == '(':
                dp[i] = dp[i - 1] + (dp[i - dp[i - 1] - 2] if i - dp[i - 1] >= 2 else 0) + 2
            max_length = max(max_length, dp[i])
    
    return max_length

# Example usage
s = "(()))())("
print(longestValidParentheses(s))  # Output: 4

Explanation

We start by making a DP array. This array helps us keep track of the longest valid parentheses lengths. We go through the string. We update the DP values by looking at the current character and the characters before it. In the end, we return the maximum value from the DP array. This value shows the length of the longest valid parentheses substring.

This dynamic programming method works well. It finds the maximum length of valid parentheses substring in O(n) time and uses O(n) space. For more information on dynamic programming, you can check this Dynamic Programming - Maximum Length of Valid Parentheses.

Dynamic Programming Approach in C++ for Maximum Length of Valid Parentheses

We will look at how to find the maximum length of a valid parentheses substring using Dynamic Programming in C++. We can use a DP array to track the lengths of valid parentheses. We will go through the string and check when we can form a valid substring.

C++ Code Implementation

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int longestValidParentheses(string s) {
    int n = s.size();
    vector<int> dp(n, 0);
    int maxLength = 0;

    for (int i = 1; i < n; i++) {
        if (s[i] == ')') {
            if (s[i - 1] == '(') {
                dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2; // Case for "()" 
            } else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(') {
                dp[i] = dp[i - 1] + (i - dp[i - 1] >= 2 ? dp[i - dp[i - 1] - 2] : 0) + 2; // Case for "()..."
            }
            maxLength = max(maxLength, dp[i]);
        }
    }
    return maxLength;
}

int main() {
    string s = "(()())";
    cout << "Maximum Length of Valid Parentheses: " << longestValidParentheses(s) << endl;
    return 0;
}

Explanation of the Code

We make a DP array dp with the same size as the input string s.

We go through each character starting from the second one. For each closing parenthesis ), we check: - If the last character is an opening parenthesis (, then we have a valid pair. - If the last character is a ), we check if there is a matching ( for the longest valid substring we found.

We will update the maxLength when we find a longer valid substring.

Complexity

• Time Complexity: O(n), where n is the length of the string.
• Space Complexity: O(n) for the DP array.

This way, we can find the maximum length of a valid parentheses substring quickly. For more info on dynamic programming methods, we can check Dynamic Programming - Maximum Length of Valid Parentheses.

Stack Based Approach for Maximum Length of Valid Parentheses in Java

To find the maximum length of valid parentheses substring using a stack-based approach in Java, we can use a stack. This stack helps us keep track of the indices of the opening parentheses. We will go through each character in the string. When we find matching closing parentheses, we can calculate the lengths of valid parentheses.

Here is how the stack-based approach works in Java:

1. Initialization: First, we create a stack to hold the indices of the opening parentheses. We also need a variable to keep track of the last index of a valid substring.

2. Iterate through the string: For each character in the string:

   • If it is an opening parenthesis (, we push its index onto the stack.
   • If it is a closing parenthesis ), we check if the stack is not empty:
     • If the stack is not empty, we pop the top of the stack and calculate the length of the valid substring.
     • If the stack is empty, we update the last index of a valid substring to the current index.

3. Calculate the maximum length: While we go through the string, we keep track of the maximum length of valid parentheses.

Here is the implementation in Java:

import java.util.Stack;

public class MaxLengthValidParentheses {
    public int longestValidParentheses(String s) {
        Stack<Integer> stack = new Stack<>();
        int maxLength = 0;
        int lastInvalidIndex = -1; // To track the last invalid index
        
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                stack.push(i); // Push the index of '('
            } else {
                if (stack.isEmpty()) {
                    lastInvalidIndex = i; // Update last invalid index
                } else {
                    stack.pop(); // Found a match for ')'
                    if (stack.isEmpty()) {
                        maxLength = Math.max(maxLength, i - lastInvalidIndex); // Calculate length
                    } else {
                        maxLength = Math.max(maxLength, i - stack.peek()); // Calculate length from the last unmatched '('
                    }
                }
            }
        }
        
        return maxLength;
    }

    public static void main(String[] args) {
        MaxLengthValidParentheses solution = new MaxLengthValidParentheses();
        String s = "(()())";
        int result = solution.longestValidParentheses(s);
        System.out.println("Maximum Length of Valid Parentheses: " + result);
    }
}

Key Points: - The stack helps us track the indices of opening parentheses easily. - The algorithm runs in O(n) time complexity. Here n is the length of the input string. - This approach calculates the maximum length of valid parentheses by updating the maximum length as we process the string.

For more information on dynamic programming and related problems, we can check the article Dynamic Programming - Longest Valid Parentheses.

Stack Based Approach for Maximum Length of Valid Parentheses in Python

We can solve the problem of finding the maximum length of valid parentheses substring using a stack in Python. Here is a simple way to do it:

1. Start a Stack: We use a stack to keep track of the indices of the parentheses. We start with a base index of -1. This helps us calculate the valid substring.

2. Go Through the String: For every character in the string:

   • If we see an opening parenthesis (, we push its index onto the stack.
   • If we see a closing parenthesis ), we check if the stack is not empty:
     • If the stack is not empty, we pop the top index from the stack. Then we calculate the length of the valid substring. We do this by subtracting the current index from the index at the new top of the stack. We update the maximum length if this new length is greater than the previous maximum.
     • If the stack is empty, we push the current index onto the stack. This helps us set a new base for future valid substrings.

3. Give Back the Maximum Length: After we go through the string, we return the maximum length we found.

Here is the Python code that shows this logic:

def maxLengthValidParentheses(s: str) -> int:
    stack = [-1]  # Start stack with base index
    max_length = 0

    for i, char in enumerate(s):
        if char == '(':
            stack.append(i)  # Push the index of '('
        else:  # char == ')'
            if stack:
                stack.pop()  # Pop the last index
                if stack:
                    # Calculate the length of valid substring
                    max_length = max(max_length, i - stack[-1])
                else:
                    # If stack is empty, push current index as base
                    stack.append(i)

    return max_length

Example Usage

s = "(()())"
print(maxLengthValidParentheses(s))  # Output: 6

s = ")()())"
print(maxLengthValidParentheses(s))  # Output: 4

s = "()(())"
print(maxLengthValidParentheses(s))  # Output: 6

This way runs in O(n) time and uses O(n) space because of the stack. It is efficient for finding the maximum length of valid parentheses substring. If you want to read more about dynamic programming, you can check this article on Dynamic Programming - Longest Valid Parentheses.

Stack Based Approach for Maximum Length of Valid Parentheses in C++

The Stack Based Approach is a simple way to find the maximum length of valid parentheses substring. This method uses a stack to track the indices of open parentheses and to find the lengths of valid substrings.

Algorithm Steps:

1. We start with an empty stack and push -1 onto it. This helps us handle the base case for valid substrings that start from index 0.
2. We look at each character in the string:
   • If we see an opening parenthesis '(', we push its index onto the stack.
   • If we see a closing parenthesis ')':
     • We pop the top element from the stack.
     • If the stack is empty after popping, we push the current index onto the stack.
     • If the stack is not empty, we find the length of the valid substring using the current index and the index at the top of the stack.
3. We keep track of the maximum length we find during the iterations.

C++ Implementation:

#include <iostream>
#include <stack>
#include <string>

using namespace std;

int maxLengthValidParentheses(string s) {
    stack<int> stk;
    stk.push(-1);
    int maxLength = 0;

    for (int i = 0; i < s.length(); i++) {
        if (s[i] == '(') {
            stk.push(i);
        } else {
            stk.pop();
            if (stk.empty()) {
                stk.push(i);
            } else {
                maxLength = max(maxLength, i - stk.top());
            }
        }
    }
    return maxLength;
}

int main() {
    string s = "(()))())(";
    cout << "Maximum Length of Valid Parentheses: " << maxLengthValidParentheses(s) << endl;
    return 0;
}

Explanation of the Code:

• The function maxLengthValidParentheses takes a string s as input.
• It starts with a stack and pushes -1 to handle the base case.
• As we go through the string, we update the stack and find the maximum length of valid parentheses based on the indices in the stack.
• The result is shown in the main function.

This stack-based approach finds the maximum length of valid parentheses quickly. It has a time complexity of O(n). So, it works well for large inputs.

Using Array for Dynamic Programming in Java

To find the maximum length of valid parentheses substring using dynamic programming in Java, we can use an array. This array will store the lengths of valid parentheses substrings that end at each index. Here is how we can do it:

1. Initialization: We need to create a DP array. This array should be the same length as the input string and we will set all values to 0.
2. Iterate Through the String: We will look at each character in the string. We need to check if it is a closing parenthesis.
3. Check for Matching Opening Parenthesis: If we find a closing parenthesis and the character before it is an opening parenthesis, we will update the DP value.
4. Update DP Array: We will use the values from the DP array to find valid substring lengths.

Here is the code to implement it:

public class MaximumLengthValidParentheses {
    public static int longestValidParentheses(String s) {
        int maxLength = 0;
        int[] dp = new int[s.length()];

        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) == ')') {
                if (s.charAt(i - 1) == '(') {
                    dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
                } else if (i - dp[i - 1] > 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
                    dp[i] = dp[i - 1] + (i >= 2 + dp[i - 1] ? dp[i - 2 - dp[i - 1]] : 0) + 2;
                }
                maxLength = Math.max(maxLength, dp[i]);
            }
        }
        return maxLength;
    }
    
    public static void main(String[] args) {
        String s = "(()())";
        System.out.println("Maximum Length of Valid Parentheses: " + longestValidParentheses(s));
    }
}

Explanation of Code:

• Dynamic Programming Array (dp): Each entry dp[i] shows the length of the longest valid parentheses substring ending at index i.
• Updating Logic:
  • If the current character is ) and the previous character is (, we have a valid pair. So, we update dp[i] to dp[i-2] + 2.
  • If the current character is ) and it forms a valid substring with a previous valid substring, we update dp[i] in another way.
• Result Calculation: We keep updating the maximum length during the loop. At the end, we return this maximum length.

This method runs in O(n) time and uses O(n) space. It is efficient for this problem. If we want to learn more about dynamic programming techniques, we can check out the Dynamic Programming on Fibonacci Numbers.

Using Array for Dynamic Programming in Python

To solve the problem of finding the longest valid parentheses substring using dynamic programming in Python, we use an array. This array helps us keep track of the lengths of valid substrings. We will go through the string and update the array based on whether the parentheses are valid.

Algorithm Steps:

1. Initialization: We create an array dp that is the same size as the string s and set all its values to zero.
2. Iterate through the string: We check each character in the string starting from the second character:
   • If the character is ')' and the one before it is '(', we set dp[i] = dp[i-2] + 2 (if i >= 2) or dp[i] = 2.
   • If the character is ')' and the one before it is also ')', we check if the character before the valid substring is '('. If it is, we set dp[i] = dp[i-1] + dp[i-dp[i-1]-2] + 2.
3. Track the maximum length: We keep a variable max_length to store the highest value we find in the dp array.

Python Code Implementation:

def longestValidParentheses(s: str) -> int:
    n = len(s)
    dp = [0] * n
    max_length = 0
    
    for i in range(1, n):
        if s[i] == ')':
            if s[i - 1] == '(':
                dp[i] = (dp[i - 2] if i >= 2 else 0) + 2
            elif i - dp[i - 1] > 0 and s[i - dp[i - 1] - 1] == '(':
                dp[i] = dp[i - 1] + (dp[i - dp[i - 1] - 2] if i - dp[i - 1] >= 2 else 0) + 2
            
            max_length = max(max_length, dp[i])
    
    return max_length

Example Usage:

s = "(()))())("
print(longestValidParentheses(s))  # Output: 4

In this example, the longest valid parentheses substring is "(())" and its length is 4.

This method runs in O(n) time and uses O(n) space. It is good for this problem. If we want to learn more about dynamic programming, we can check out Dynamic Programming - Unique Paths in a Grid.

Frequently Asked Questions

1. What is the dynamic programming approach to finding the maximum length of valid parentheses?

We use dynamic programming to solve problems by breaking them into easier parts. For the maximum length of valid parentheses, we create a DP array. Each spot in this array shows the maximum length of valid parentheses that ends at that position. We go through the string and use simple rules based on the character. This way, we fill the array and find the maximum valid length quickly.

2. How does the stack-based approach work for valid parentheses?

We can use a stack to find the maximum length of valid parentheses. The stack keeps track of the positions of opening parentheses. When we see a closing parenthesis, we check if there is a matching opening one in the stack. If we find it, we calculate the length of the valid substring. Then, we update the maximum length. This method helps us handle nested and consecutive valid parentheses easily.

3. Can you explain the time and space complexity of the maximum length of valid parentheses problem?

The time complexity for both the dynamic programming and stack-based methods is O(n). Here, n is the length of the input string. We need to go through the string one time. The space complexity for the stack method is O(n) because we store indices. The dynamic programming method also needs O(n) space for the DP array.

4. What are some common variations of the valid parentheses problem?

Some common variations of the valid parentheses problem are finding the longest valid parentheses substring, counting all valid parentheses combinations, and checking if a string of parentheses is valid. We can use similar dynamic programming or stack methods for these variations. They are good practice for improving our problem-solving skills in competitive programming.

5. Where can I find more examples of dynamic programming problems like valid parentheses?

For more examples of dynamic programming problems, we can check out articles like Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs. These resources give clear explanations and code examples to help us learn dynamic programming techniques.