Minimum Remove to Make Valid Parentheses | 3 Detailed Approaches | Leetcode 1249 | codestorywithMIK
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This is the 15th Video of our Playlist "STACK : Popular Interview Problems".
In this video we will try to solve a pretty famous stack based Problem :
Minimum Remove to Make Valid Parentheses | 3 Detailed Approaches | Leetcode 1249 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Minimum Remove to Make Valid Parentheses | 3 Detailed Approaches | Leetcode 1249 | codestorywithMIK
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My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/strings/Minimum%20Remove%20to%20Make%20Valid%20Parentheses.cpp
Leetcode Link : https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/
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APPROACHES SUMMARY :
Here's a short summary of each approach to solve the problem of removing the minimum number of parentheses to make a string valid:
### Approach-1 (Using Stack and Set)
1. Utilizes a stack to keep track of open parentheses '('.
2. If encountering a closing parenthesis ')', checks if the stack is empty. If so, marks the current index for removal. Otherwise, pops the last open parenthesis from the stack, indicating a valid pair.
3. After iterating through the string, any remaining open parentheses in the stack are marked for removal.
4. Finally, constructs the resulting string excluding the marked indices.
### Approach-2 (Iterating Front and Back to Eliminate)
1. Iterates through the string from the front, eliminating invalid parentheses by keeping track of the count of open parentheses encountered.
2. Repeats the process from the back to eliminate invalid closing parentheses.
3. Utilizes two passes through the string to ensure validity, and reverses the resulting string obtained from the back iteration.
### Approach-3 (Simplified and Straightforward)
1. Iterates through the string once, keeping track of the count of open parentheses encountered.
2. Constructs a temporary string with valid parentheses, excluding any excessive closing parentheses.
3. Reverses the temporary string, removing any excessive open parentheses from the end.
4. Returns the final resulting string.
Each approach provides a different perspective on solving the problem, with varying degrees of complexity and efficiency.
✨ Timelines✨
00:00 - Introduction
03:18 - Approach-1 Intuition
11:55 - Coding Approach-1
15:44 - Approach-2 Intuition
29:07 - Coding Approach-2
35:07 - Approach-3 Intuition
39:50 - Coding Approach-3
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