[Dynamic Programming] Longest Valid Parentheses - Medium

The Longest Valid Parentheses problem is a well-known challenge in dynamic programming. It is about finding the length of the longest valid parentheses substring in a string of parentheses. A valid substring has every opening parenthesis matched with a closing one. They must be in the right order. Our goal is to find this length in an efficient way using dynamic programming or stack methods.

In this article, we will look at different ways to solve the Longest Valid Parentheses problem. First, we will explain the problem. Next, we will discuss dynamic programming methods in Java, Python, and C++. After that, we will check out optimized stack methods in these same languages. We will also compare the dynamic programming and stack methods. Plus, we will give a detailed code walkthrough for the Java version. Lastly, we will answer common questions about this topic.

  • [Dynamic Programming] Longest Valid Parentheses Problem Explained
  • Dynamic Programming Approach in Java
  • Dynamic Programming Approach in Python
  • Dynamic Programming Approach in C++
  • Optimized Stack Approach in Java
  • Optimized Stack Approach in Python
  • Optimized Stack Approach in C++
  • Dynamic Programming vs Stack Approach Comparison
  • Code Walkthrough for Java Implementation
  • Frequently Asked Questions

For more insights into dynamic programming, we can check articles on the Fibonacci Number and Climbing Stairs. They are very helpful.

Dynamic Programming Approach in Java

To solve the Longest Valid Parentheses problem with dynamic programming in Java, we can make a DP array. In this array, dp[i] shows the length of the longest valid parentheses substring that ends at index i.

Steps:

  1. First, we create a DP array that is the same length as the input string. We fill it with zeros.
  2. Then, we use a loop to go through the string. For every closing parenthesis ')', we check if it can make a valid substring with a previous '('.
  3. We update the DP array based on these rules:
    • If s[i] is ')' and s[i-1] is '(', we set dp[i] = dp[i-2] + 2.
    • If s[i] is ')' and s[i-1] is ')', we need to check if s[i - dp[i-1] - 1] is '('. If it is, we set dp[i] = dp[i-1] + 2 + dp[i - dp[i-1] - 2].

Java Code Implementation:

public class LongestValidParentheses {
    public int longestValidParentheses(String s) {
        if (s == null || s.length() == 0) return 0;

        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] - 1 >= 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
                    dp[i] = dp[i - 1] + 2 + (i - dp[i - 1] - 2 >= 0 ? dp[i - dp[i - 1] - 2] : 0);
                }
                maxLength = Math.max(maxLength, dp[i]);
            }
        }
        return maxLength;
    }

    public static void main(String[] args) {
        LongestValidParentheses lvp = new LongestValidParentheses();
        String s = "(()))())(";
        System.out.println("Longest Valid Parentheses Length: " + lvp.longestValidParentheses(s)); // Output: 4
    }
}

Key Properties:

  • Time Complexity: O(n). Here n is the length of the string.
  • Space Complexity: O(n) for the DP array.
  • This method works well to find the length of the longest valid parentheses substring using dynamic programming ideas.

This way is strong for many inputs. It can handle edge cases like empty strings or strings without valid parentheses.

Dynamic Programming Approach in Python

To solve the Longest Valid Parentheses problem with dynamic programming in Python, we create a DP array. Here, dp[i] shows the length of the longest valid parentheses substring that ends at index i.

Steps:

  1. First, we make a DP array of size n (the length of the string) and fill it with zeros.
  2. Next, we go through the string starting from index 1.
  3. For each character, we check if it makes a valid pair with the last one or if it helps to form a longer valid substring.
  4. Then we update the DP array based on the rules for valid parentheses.

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] == '(':  # Case of "..."
                dp[i] = (dp[i - 2] if i >= 2 else 0) + 2
            elif i - dp[i - 1] > 0 and s[i - dp[i - 1] - 1] == '(':  # Case of ")..."
                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: 6

Explanation:

  • The code goes through the string and looks for valid pairs of parentheses.
  • It updates the dp array to track valid lengths. At the end, it returns the maximum length.
  • Complexity: O(n) time and O(n) space, where n is the length of the input string.

This way is good to find the longest valid parentheses substring using dynamic programming ideas. It works well and fast. For more information on dynamic programming, we can check Dynamic Programming - Fibonacci Number and other similar topics.

Dynamic Programming Approach in C++

We can solve the Longest Valid Parentheses problem by using a dynamic programming method in C++. The main idea is to keep a dp array. In this array, dp[i] shows the length of the longest valid parentheses substring that ends at index i.

Algorithm Steps:

  1. First, we make a dp array. The size of this array is equal to the length of the input string. We fill it with zeros.
  2. Then, we go through the string starting from the second character.
  3. For each character:
    • If the current character is ')', we check the character before it:
      • If the character before it is '(', then we set dp[i] = dp[i-2] + 2.
      • If the character before it is ')', we check if the character before the valid matching parenthesis is '(':
        • If it is, we set dp[i] = dp[i-1] + 2 + dp[i - dp[i-1] - 2] if it exists.
  4. We also keep a maxLength variable. This variable stores the biggest value we find in the dp array.

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] + 2 : 2;
            } else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(') {
                dp[i] = dp[i - 1] + 2 + ((i - dp[i - 1] - 2 >= 0) ? dp[i - dp[i - 1] - 2] : 0);
            }
            maxLength = max(maxLength, dp[i]);
        }
    }
    return maxLength;
}

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

Time Complexity:

  • O(n): We look at the string one time.

Space Complexity:

  • O(n): We use a dp array to keep the lengths.

This way, we can find the longest valid parentheses substring using dynamic programming in C++. For more reading about similar dynamic programming problems, we can check out Dynamic Programming: Fibonacci Number.

Optimized Stack Approach in Java

We can use the Optimized Stack Approach to solve the Longest Valid Parentheses problem. This method uses a stack to track the indices of unmatched parentheses. It helps us find the longest substring of valid parentheses quickly. We can do this in O(n) time and O(n) space.

Algorithm Steps:

  1. First, we need to set up a stack to hold the indices of parentheses.
  2. Then, we push -1 onto the stack. This helps us handle edge cases.
  3. Next, we go through the string:
    • If we see ‘(’, we push its index onto the stack.
    • If we see ‘)’, we pop the top of the stack:
      • If the stack is not empty, we find the length of the valid substring. We do this by subtracting the index at the top of the stack from the current index.
      • If the stack is empty, we push the current index onto the stack.
  4. We also keep track of the maximum length of valid parentheses we find while we go through the string.

Java Code Implementation:

public class LongestValidParentheses {
    public int longestValidParentheses(String s) {
        Stack<Integer> stack = new Stack<>();
        stack.push(-1);
        int maxLength = 0;

        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                stack.push(i);
            } else {
                stack.pop();
                if (!stack.isEmpty()) {
                    maxLength = Math.max(maxLength, i - stack.peek());
                } else {
                    stack.push(i);
                }
            }
        }

        return maxLength;
    }

    public static void main(String[] args) {
        LongestValidParentheses lvp = new LongestValidParentheses();
        String input = "(()())";
        System.out.println("Longest Valid Parentheses Length: " + lvp.longestValidParentheses(input)); // Output: 6
    }
}

Key Points:

  • This way is good for large strings with parentheses.
  • We use a stack to manage indices well.
  • We update the maximum length of valid parentheses as we check each character.

This method works great for problems with valid parentheses. We can also use it for similar challenges in dynamic programming. For more on dynamic programming, we can check Dynamic Programming - Fibonacci Number.

Optimized Stack Approach in Python

We can use the Optimized Stack Approach to solve the Longest Valid Parentheses problem. This method uses a stack to keep track of the positions of opening parentheses. It helps us find a solution in linear time, which is the best for this problem.

Algorithm Steps:

  1. First, we set up a stack and add -1 to it. This helps us with edge cases.
  2. Next, we go through each character in the string:
    • If we see ‘(’, we add its index to the stack.
    • If we see ‘)’:
      • We take the top index from the stack.
      • If the stack is not empty, we find the length of the valid substring. We do this by subtracting the top index from the current index.
      • If the stack is empty, we add the current index to the stack.
  3. We also keep track of the longest valid parentheses we find while going through the string.

Python Implementation:

def longestValidParentheses(s: str) -> int:
    stack = [-1]
    max_length = 0

    for i, char in enumerate(s):
        if char == '(':
            stack.append(i)
        else:
            stack.pop()
            if stack:
                max_length = max(max_length, i - stack[-1])
            else:
                stack.append(i)

    return max_length

# Example usage
input_str = "(()())"
print(longestValidParentheses(input_str))  # Output: 6

Explanation of the Code:

  • The longestValidParentheses function takes a string s as input.
  • We start the stack with -1 to help calculate lengths right.
  • We loop through each character. We manage the stack based on if the character is an opening or closing parenthesis.
  • When we find a valid pair, we update the maximum length.

This optimized stack approach helps us find the length of the longest valid parentheses substring. It does this with a time complexity of O(n) and space complexity of O(n). If you want to learn more about similar topics in dynamic programming, you can check the Dynamic Programming: Longest Common Subsequence.

Optimized Stack Approach in C++

The Optimized Stack Approach helps us solve the Longest Valid Parentheses problem in C++. We use a stack to track the indices of the parentheses. This method processes the input string well. It helps us find the length of the longest valid substring of parentheses.

Algorithm Steps:

  1. We start by making a stack that will hold the indices of opening parentheses.
  2. We push -1 onto the stack. This helps with cases where valid parentheses begin at the start.
  3. We go through each character in the string:
    • If we find an opening parenthesis '(', we push its index onto the stack.
    • If we find a closing parenthesis ')', we do the following:
      • We pop the top of the stack. This gives us the index of the last unmatched opening parenthesis.
      • If the stack is not empty after popping, we find the length of the valid substring. We use the current index minus the index at the new top of the stack.
      • If the stack is empty, we push the current index onto the stack.
  4. We keep track of the maximum length of valid parentheses we find while going through the string.

C++ Implementation:

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

int longestValidParentheses(std::string s) {
    std::stack<int> stack;
    stack.push(-1); // Base for valid parentheses
    int maxLength = 0;

    for (int i = 0; i < s.length(); i++) {
        if (s[i] == '(') {
            stack.push(i); // Push index of '('
        } else {
            stack.pop(); // Pop the last '(' index
            if (!stack.empty()) {
                // Current valid substring length
                maxLength = std::max(maxLength, i - stack.top());
            } else {
                stack.push(i); // Push current index for future valid lengths
            }
        }
    }
    return maxLength;
}

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

Complexity Analysis:

  • Time Complexity: O(n). Here n is the length of the string. Each character is checked once.
  • Space Complexity: O(n) in the worst case for the stack. But it could also be O(1) if we only use simple variables to track indices.

This optimized stack approach works well for the Longest Valid Parentheses problem. It is clear and efficient. If you want to read more about similar dynamic programming problems, check out the Dynamic Programming - Fibonacci Number.

Dynamic Programming vs Stack Approach Comparison

When we solve the Longest Valid Parentheses problem, we can use two common methods. These are the Dynamic Programming approach and the Stack approach. Each way has its good points and some downsides.

Dynamic Programming Approach

  • Time Complexity: O(n)
  • Space Complexity: O(n)
  • How it Works:
    • We keep a DP array. Here, dp[i] shows the length of the longest valid parentheses substring that ends at index i.
    • We go through the string and update the DP array based on the rules for valid parentheses.

Example Code in Java:

public int longestValidParentheses(String s) {
    int maxLen = 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;
            }
            maxLen = Math.max(maxLen, dp[i]);
        }
    }
    return maxLen;
}

Stack Approach

  • Time Complexity: O(n)
  • Space Complexity: O(n)
  • How it Works:
    • We use a stack to keep track of the indices of the parentheses.
    • We push the index of ( onto the stack. When we see ), we pop the stack and find the length of valid parentheses from the current position and the top of the stack.

Example Code in Java:

public int longestValidParentheses(String s) {
    Stack<Integer> stack = new Stack<>();
    stack.push(-1);
    int maxLen = 0;

    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == '(') {
            stack.push(i);
        } else {
            stack.pop();
            if (stack.isEmpty()) {
                stack.push(i);
            } else {
                maxLen = Math.max(maxLen, i - stack.peek());
            }
        }
    }
    return maxLen;
}

Key Differences

  • Data Structure: Dynamic Programming uses an array to keep results. The Stack approach uses a stack to manage indices.
  • Memory Usage: DP can use more memory than the stack approach. It depends on the size of the input.
  • Implementation Complexity: The stack method can be easier to implement for this problem.

Both ways are good for solving the Longest Valid Parentheses problem. But the choice can depend on the specific situation or limits of the problem we solve. For more reading on dynamic programming techniques, we can look at the Dynamic Programming - Fibonacci Number article.

Code Walkthrough for Java Implementation

In this section, we will look at how to solve the Longest Valid Parentheses problem in Java. We will use a dynamic programming method. The goal is to find the length of the longest valid parentheses substring.

Java Implementation

Here is the Java code for the Longest Valid Parentheses problem:

public class LongestValidParentheses {
    public 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) == '(') {
                    // Case: "..."
                    dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
                } else if (i - dp[i - 1] > 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
                    // Case: "..."
                    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) {
        LongestValidParentheses lvp = new LongestValidParentheses();
        String s = "(()))())(";
        System.out.println("Longest Valid Parentheses Length: " + lvp.longestValidParentheses(s));
    }
}

Explanation of the Code

  • Dynamic Programming Array (dp): We use an array dp. Here, dp[i] keeps the length of the longest valid parentheses substring that ends at index i.

  • Iterating through the string: We start from index 1. For each closing parenthesis ), we check:

    • If it comes after an opening parenthesis (, we can make a valid pair. We add 2 to the length of the valid substring before these two parentheses.
    • If it comes after a valid substring, we see if we can add to it by looking at the character before the valid substring.

  • Updating maxLength: After we check each character, we update maxLength with the biggest value found in dp.

Example

For the input string s = "(()))())(", the output will be:

Longest Valid Parentheses Length: 4

This shows that the longest valid parentheses substring is "(())".

This code works well to find the longest valid parentheses using dynamic programming. It runs in O(n) time, where n is the length of the input string s. For more about dynamic programming, we can look at related problems like Dynamic Programming - Fibonacci Number or Dynamic Programming - Climbing Stairs.

Frequently Asked Questions

1. What is the Longest Valid Parentheses problem?

The Longest Valid Parentheses problem is a popular coding challenge. We need to find the length of the longest valid parentheses substring in a string of parentheses. We can solve this problem using dynamic programming or a stack. This helps us practice algorithm skills and understand how parentheses match.

2. How does dynamic programming solve the Longest Valid Parentheses problem?

Dynamic programming solves the Longest Valid Parentheses problem by splitting it into smaller problems. It uses a DP array to keep track of the longest valid substring that ends at each index. We go through the string and update the DP array based on the current character and its previous states. This way, we can find the length of the longest valid parentheses substring quickly.

3. What is the stack approach for the Longest Valid Parentheses problem?

The stack approach for the Longest Valid Parentheses problem uses a stack to track the indices of unmatched parentheses. We push the index of each opening parenthesis. For closing ones, we pop from the stack. This helps us find the lengths of valid segments. This method is fast with O(n) complexity, which is good for this problem.

4. How do I implement the Longest Valid Parentheses solution in Python?

To implement the Longest Valid Parentheses solution in Python, we can use dynamic programming or a stack. The stack method is usually easier to understand. Here is a quick code snippet for the stack approach:

def longestValidParentheses(s: str) -> int:
    stack = [-1]
    max_length = 0
    for i, char in enumerate(s):
        if char == '(':
            stack.append(i)
        else:
            stack.pop()
            if not stack:
                stack.append(i)
            else:
                max_length = max(max_length, i - stack[-1])
    return max_length

This code finds the length of the longest valid parentheses substring quickly.

5. What are the differences between dynamic programming and stack approaches for this problem?

The dynamic programming approach for the Longest Valid Parentheses problem builds solutions from smaller parts. It uses a DP array to manage states. The stack approach uses a stack to track indices and find valid lengths directly. Both methods have O(n) time complexity. But the stack method often uses less space and is easier to understand for this problem.

For more reading on dynamic programming ideas, check out articles like Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs.