2134. Minimum Swaps to Group All 1's Together II
Medium
A swap is defined as taking two distinct positions in an array and swapping the values in them.
A circular array is defined as an array where we consider the first element and the last element to be adjacent.
Given a binary circular array nums, return the minimum number of swaps required to group all 1's present in the array together at any location.
Example 1:
- Input: nums = [0,1,0,1,1,0,0]
- Output: 1
-
Explanation: Here are a few of the ways to group all the 1's together:
- [0,0,1,1,1,0,0] using 1 swap.
- [0,1,1,1,0,0,0] using 1 swap.
- [1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).
- There is no way to group all 1's together with 0 swaps.
- Thus, the minimum number of swaps required is 1.
Example 2:
- Input: nums = [0,1,1,1,0,0,1,1,0]
- Output: 2
-
Explanation: Here are a few of the ways to group all the 1's together:
- [1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).
- [1,1,1,1,1,0,0,0,0] using 2 swaps.
- There is no way to group all 1's together with 0 or 1 swaps.
- Thus, the minimum number of swaps required is 2.
Example 3:
- Input: nums = [1,1,0,0,1]
- Output: 0
-
Explanation: All the 1's are already grouped together due to the circular property of the array.
- Thus, the minimum number of swaps required is 0.
Constraints:
- 1 5
- nums[i] is either 0 or 1.
Hint:
- Notice that the number of 1’s to be grouped together is fixed. It is the number of 1's the whole array has.
- Call this number total. We should then check for every subarray of size total (possibly wrapped around), how many swaps are required to have the subarray be all 1’s.
- The number of swaps required is the number of 0’s in the subarray.
- To eliminate the circular property of the array, we can append the original array to itself. Then, we check each subarray of length total.
- How do we avoid recounting the number of 0’s in the subarray each time? The Sliding Window technique can help.
Solution:
To solve this problem, we can follow these steps:
- Count the Total Number of 1s: This will be the number of 1s we need to group together.
- Extend the Array: To handle the circular nature, append the array to itself.
- Use Sliding Window Technique: Apply the sliding window technique on the extended array to find the minimum number of swaps required.
Let's implement this solution in PHP: 2134. Minimum Swaps to Group All 1's Together II
<?php // Example usage $nums1 = [0,1,0,1,1,0,0]; $nums2 = [0,1,1,1,0,0,1,1,0]; $nums3 = [1,1,0,0,1]; echo minSwaps($nums1) . "\n"; // Output: 1 echo minSwaps($nums2) . "\n"; // Output: 2 echo minSwaps($nums3) . "\n"; // Output: 0 ?>
Explanation:
- Count the Total Number of 1s: Calculate the total number of 1s in the original array.
- Extend the Array: Concatenate the original array to itself to handle the circular nature.
- Initial Window: Count the number of 0s in the initial window of size equal to the total number of 1s.
- Sliding Window: Slide the window across the extended array. For each new position, update the count of 0s based on the elements entering and leaving the window.
- Find Minimum: Keep track of the minimum number of 0s encountered, which corresponds to the minimum number of swaps needed.
This solution efficiently handles the circular array by transforming it into a linear problem and uses the sliding window technique to maintain a running count of 0s in each window of size equal to the total number of 1s.
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