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Longest Valid Parentheses

Certification: Advanced Level Accuracy: 100% Submissions: 1 Points: 15

Write a Java program to find the length of the longest valid parentheses substring. A valid parentheses substring is a string of parentheses where every opening parenthesis '(' has a matching closing parenthesis ')' and they are correctly nested.

Example 1
    
  
  • Input: s = "(()"
    • 
  
  • Output: 2
    • 
  
  • Explanation: The last two characters "()" form a valid parentheses pair.
    The length of this valid substring is 2.
    Therefore, the longest valid parentheses substring has length 2.
    •
Example 2
    
  
  • Input: s = ")()())"
    • 
  
  • Output: 4
    • 
  
  • Explanation: The first character ")" has no matching opening parenthesis, so it's invalid.
    The next four characters "()()" form a valid parentheses substring.
    The length of this valid substring is 4.
    Therefore, the longest valid parentheses substring has length 4.
    •
Constraints
    
  
  • 0 <= s.length <= 3 * 10^4
    • 
  
  • s consists of only '(' and ')' characters
    • 
  
  • Time Complexity: O(n) where n is the length of the string
    • 
  
  • Space Complexity: O(n)
    •
StringsStackDropboxD. E. Shaw
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Hint
Solution

Solution Hints

    
  
  • Use a stack data structure to keep track of the indices of opening parentheses
    • 
  
  • Push -1 onto the stack initially as a base marker
    • 
  
  • Iterate through the string, if the current character is '(', push its index onto the stack
    • 
  
  • If the current character is ')', pop the top element from the stack
    • 
  
  • If the stack becomes empty, push the current index as a new base marker
    • 
  
  • If the stack is not empty, calculate the length of the current valid substring (current index - index at the top of the stack)
    • 
  
  • Update the maximum length found so far
    •

Steps to solve by this approach:

 Step 1: Initialize a stack and push -1 as the base index to calculate substring lengths.

 Step 2: Iterate through each character in the string.

 Step 3: If the current character is '(', push its index onto the stack.

 Step 4: If the current character is ')', pop from the stack.

 Step 5: If the stack becomes empty after popping, push the current index as a new base marker.

 Step 6: Otherwise, calculate the length of the current valid substring (current index - index at top of stack).

 Step 7: Return the maximum valid substring length found.

Submitted Code :