LEETCODE 1249 C SOLUTION:efficiently Remove Invalid Parentheses to Make a String Valid

Welcome to this comprehensive tutorial on solving LeetCode Problem 1249: "Remove Invalid Parentheses". In this video, we'll dive deep into understanding the problem statement, explore various approaches to solve it, and implement an efficient solution using the stack method. This video is perfect for those looking to enhance their problem-solving skills and prepare for coding interviews.
Problem Statement:
The problem requires us to remove the minimum number of parentheses from a given string to make it valid. A valid string is one where every open parenthesis '(' has a corresponding closing parenthesis ')', and the pairs are properly nested.
Example:
Let's consider the string s = "lee(t(c)o)de)". This string has an extra closing parenthesis at the end. The goal is to remove the minimum number of parentheses to make it valid. The expected result would be "lee(t(c)o)de".
Approach:
To solve this problem, we'll use a stack-based approach. This method ensures that we efficiently track unmatched parentheses and remove the invalid ones in a systematic way. Here's a detailed step-by-step breakdown of our approach:
Initialization:
We'll use a stack to store the indices of the open parentheses '(' that don't have a matching closing parenthesis ')'.
We'll iterate through the string and mark the invalid parentheses.
First Pass - Marking Invalid Parentheses:
Traverse the string and for each character:
If it's an open parenthesis '(', push its index onto the stack.
If it's a closing parenthesis ')', check if the stack is empty:
If the stack is empty, mark this closing parenthesis as invalid by replacing it with a placeholder (e.g., '*').
If the stack is not empty, pop from the stack, indicating that we found a matching pair.
Second Pass - Mark Remaining Unmatched Open Parentheses:
After the first pass, any indices left in the stack correspond to unmatched open parentheses '('.
Mark these positions in the string as invalid by replacing them with the placeholder '*'.
Removing Invalid Parentheses:
Use a method to remove all placeholder characters from the string.
The resulting string is the valid string with the minimum number of parentheses removed.
Benefits of This Approach:
Efficiency: The stack method ensures we only traverse the string a couple of times, making it efficient with a linear time complexity of O(n), where n is the length of the string.
Simplicity: This method is straightforward to implement and understand, making it a great choice for interview preparation.
Example Walkthrough:
Let's walk through an example to see how our approach works in practice.
Consider the input string: s = "(a(b(c)d)".
First Pass:
Initialize an empty stack.
Traverse the string:
Encounter '(': push its index (0) onto the stack.
Encounter 'a': continue.
Encounter '(': push its index (2) onto the stack.
Encounter 'b': continue.
Encounter '(': push its index (4) onto the stack.
Encounter 'c': continue.
Encounter ')': pop from the stack (index 4), found a pair.
Encounter 'd': continue.
Encounter ')': pop from the stack (index 2), found a pair.
Stack now contains index [0], indicating an unmatched '(' at index 0.
Second Pass:
Mark the unmatched '(' at index 0 with a placeholder '*'.
The modified string becomes "*a(b(c)d)".
Remove Placeholders:
Remove all '*' characters.
The final valid string is "a(b(c)d)".
Conclusion:
By the end of this video, you'll have a solid understanding of how to approach and solve the "Remove Invalid Parentheses" problem using the stack method. We'll also discuss potential edge cases and how to handle them effectively.
Summary:
Problem Statement: Remove the minimum number of parentheses to make a string valid.
Approach: Use a stack to track unmatched parentheses and mark invalid ones.
Steps:
Traverse the string and mark invalid parentheses.
Mark remaining unmatched open parentheses.
Remove placeholders to get the valid string.
Efficiency: Linear time complexity O(n).
Additional Tips:
Edge Cases: Consider strings with no parentheses, strings where all parentheses are already valid, and strings with only one type of parenthesis.
Debugging: Use print statements to track the indices stored in the stack and the modifications made to the string at each step.
Practice: Try solving similar problems on LeetCode to reinforce your understanding.
Thank you for watching this tutorial. If you found this video helpful, please like, subscribe, and share it with others preparing for coding interviews. Feel free to leave any questions or comments below, and I'll be happy to help. Happy coding!