Minimum Swaps to Group All 1's Together II | 2 Ways | Dry Runs | Leetcode 2134 | codestorywithMIK
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This is the 25th Video of our Playlist "Sliding Window : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good and standard Sliding Window Problem : Minimum Swaps to Group All 1's Together II | 2 Ways | Dry Runs | Leetcode 2134 | 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 Swaps to Group All 1's Together II | 2 Ways | Dry Runs | Leetcode 2134 | codestorywithMIK
Company Tags : Microsoft
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Summary :
Approach 1: Using Extra Space and Sliding Window
Time Complexity: O(n)
Space Complexity: O(2n) ~ O(n)
Explanation:
Duplicating the Array: A temporary array temp of size 2*n is created by concatenating the original array with itself. This allows for a circular array approach using a linear array.
Count of 1s: The total number of 1s in the original array is counted using accumulate.
Sliding Window: A sliding window of size equal to the number of 1s is used to find the maximum number of 1s in any subarray of this size. This is done by:
Initializing pointers i and j to the start of the array.
Moving j to expand the window and counting 1s.
If the window size exceeds the number of 1s, moving i to shrink the window and adjusting the count of 1s.
Tracking the maximum number of 1s found in any window of the required size.
Result Calculation: The minimum swaps needed is calculated as the total number of 1s minus the maximum number of 1s found in any window.
Approach 2: Sliding Window without Extra Space
Time Complexity: O(n)
Space Complexity: O(1)
Explanation:
Count of 1s: Similar to Approach 1, the total number of 1s in the original array is counted.
Sliding Window with Modulo Operation: Instead of duplicating the array, the modulo operation (j % n and i % n) is used to handle the circular nature of the array within the same array space.
Sliding Window: The logic for the sliding window remains the same as in Approach 1:
Using pointers i and j to define the window.
Adjusting the count of 1s when expanding or shrinking the window.
Tracking the maximum number of 1s found.
Result Calculation: The result is calculated as the total number of 1s minus the maximum number of 1s found in any window.
Comparison:
Both approaches have the same time complexity of O(n).
Approach 1 uses additional space to create a duplicated array, resulting in O(n) space complexity.
Approach 2 optimizes space usage by leveraging the modulo operation, resulting in O(1) space complexity.
✨ Timelines✨
00:00 - Introduction
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