Minimum Number of Swaps to Make the String Balanced | Reason | Leetcode 1963 | codestorywithMIK
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This is the 23rd Video of our Playlist "Stack : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve an easy stack based Problem : Minimum Number of Swaps to Make the String Balanced | Detailed Reason | Leetcode 1963 | 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 Number of Swaps to Make the String Balanced | Detailed Reason | Leetcode 1963 | codestorywithMIK
Company Tags : will update soon
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Stack/Minimum Number of Swaps to Make the String Balanced.cpp
Leetcode Link : https://leetcode.com/problems/minimum-number-of-swaps-to-make-the-string-balanced/
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Summary :
Approach 1: Using Stack
Time Complexity: O(n)
Space Complexity: O(n)
This approach uses a stack to keep track of unmatched opening brackets ([). When a closing bracket (]) is encountered, it checks if the stack is not empty, indicating a corresponding unmatched opening bracket exists. If so, it pops the stack, effectively pairing the brackets. At the end of the traversal, the remaining elements in the stack indicate unmatched opening brackets. The minimum number of swaps required to balance the string is calculated using the formula (stack.size() + 1) / 2.
Pros: Simple to understand, directly simulates bracket pairing.
Cons: Uses extra space proportional to the number of unmatched brackets.
Approach 2: Without Using Stack
Time Complexity: O(n)
Space Complexity: O(1)
Instead of using a stack, this approach uses a counter (size) to keep track of unmatched opening brackets. For every [ encountered, the counter is incremented, and for each ] encountered, it is decremented if the counter is non-zero (indicating an unmatched opening bracket is available). After processing the entire string, the remaining value in size represents the number of unmatched opening brackets. The number of swaps is calculated as (size + 1) / 2.
✨ Timelines✨
00:00 - Introduction
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