The problem of finding the longest valid parentheses in a string is a common algorithmic challenge that can be approached by understanding some fundamental principles of string processing and stack data structures. To break it down from first principles:

Basic Concepts:

1. String Processing:

   • A string is a sequence of characters.
   • In this problem, we're interested in two specific characters: opening parentheses '(' and closing parentheses ')'.
   • A valid parentheses sequence is one where every opening parenthesis '(' is matched with a closing parenthesis ')' in the correct order.

2. Stack Data Structure:

   • A stack is a linear data structure which follows the Last In, First Out (LIFO) principle.
   • It has two primary operations: push, which adds an element to the top of the stack, and pop, which removes the top element.
   • Stacks are ideal for keeping track of order and pairing, which is crucial in problems involving balanced parentheses.

Problem Understanding:

• Goal: Find the length of the longest substring of a given string that is a "valid" parentheses string.
• Valid Parentheses String: A string of parentheses is considered valid if every '(' has a corresponding ')' and they are correctly ordered. For example, "()" and "()()" are valid, but ")(" and "(()" are not.

Solution Approach:

1. Using a Stack:

   • Initialize a stack and push -1 onto it. This serves as a base for the first valid string.
   • Iterate through each character in the string.
   • If the character is '(':
     • Push its index onto the stack.
   • If the character is ')':
     • Pop the top element from the stack.
     • If the stack is empty after this pop, push the current index onto the stack.
     • If the stack is not empty, calculate the length of the current valid string by subtracting the current index with the top element of the stack.
     • Update the longest length if the current length is greater.

2. Dynamic Programming:

   • Use an array to store the length of the longest valid parentheses ending at each index.
   • Iterate through the string:
     • If a character is '(': Skip, as no valid string can end with '('.
     • If a character is ')': Check the character before it.
       • If it is '(', then a pair is formed. Add 2 to the length at the index two positions before.
       • If it is ')', check if it forms a valid sequence with a previous '('. This is determined by the value in the DP array at the index corresponding to the position before the previous pair.

Complexity:

• Time Complexity:
  • Both approaches have a time complexity of O(n), where n is the length of the string.
• Space Complexity:
  • Using a stack requires O(n) space.
  • The dynamic programming approach also uses O(n) space for the array.

Implementation:

Implementing these algorithms involves careful index management and understanding of how stacks and dynamic programming arrays store information. The stack approach is often more intuitive, as it directly simulates the process of pairing parentheses. The dynamic programming approach, while more complex in understanding, eliminates the need for an explicit stack and can be more efficient in terms of constant factors.