Python | Sliding Window
see the Successfully Accepted Submission\n\n# Code\n\nclass Solution:\n    def minSwaps(self, nums: List[int]) -> int:\n        n = len(nums)\n        k = sum(n
[see the Successfully Accepted Submission](https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together-ii/submissions/1341619703/?envType=daily-question&envId=2024-08-02)\n\n# Code\n```\nclass Solution:\n    def minSwaps(self, nums: List[int]) -> int:\n        n = len(nums)\n        k = sum(nums) \n        if k == 0 or k == n:\n            return 0  \n\n        extended_nums = nums + nums[:k-1]\n        current_zeros = sum(1 for i in range(k) if extended_nums[i] == 0)\n        min_zeros = current_zeros\n        \n        for i in range(1, n):\n            if extended_nums[i-1] == 0:\n                current_zeros -= 1\n            if extended_nums[i+k-1] == 0:\n                current_zeros += 1\n            min_zeros = min(min_zeros, current_zeros)\n        \n        return min_zeros\n        \n```\n\n\n## Steps:\n\n### 1. Count the Total Number of 1s:\n- First, we need to count the total number of 1s in the array. Let\'s call this count `k`.\n\n### 2. Initial Window Setup:\n- We\'ll use a sliding window of size `k` to traverse the array. This window will help us count the number of 0s in any window of size `k`.\n\n### 3. Calculate Initial Number of 0s in the First Window:\n- Set up the initial window and count how many 0s are in this window.\n\n### 4. Slide the Window:\n- Move the window one position at a time. For each move, adjust the count of 0s based on the elements that are entering and leaving the window.\n- Track the minimum number of 0s found in any window of size `k`.\n\n### 5. Handle Circular Nature:\n- Since the array is circular, extend the array by appending the first `k-1` elements at the end of the array. This allows the sliding window to naturally wrap around the end of the array.\n\n### 6. Result:\n- The result will be the minimum number of 0s found in any window of size `k`, as this represents the fewest swaps needed to group all 1s together.\n\n## Explanation:\n\n### Counting 1s (`k`):\n- We count the total number of 1s to determine the window size.\n\n### Sliding Window:\n- By extending the array, we handle the circular nature seamlessly.\n\n### Tracking Minimum 0s:\n- As we slide the window, we keep track of the minimum number of 0s within any window of size `k`, which corresponds to the minimum swaps needed.\n\n\n\n![image](https://assets.leetcode.com/users/images/b1e4af48-4da1-486c-bc97-b11ae02febd0_1696186577.992237.jpeg)
['Python3']