O(N) || EASY- EXPLAINED ||JAVA C++ PYTHON
HEY , IF THE HELPS YOU THEN PLS UPVOTE IT . THANKS @aryan43 for C++ code. Hope to see your knight badge soon.\n\n### Approach\n\n1. Count the total number of 1s
# HEY , IF THE HELPS YOU THEN PLS UPVOTE IT . THANKS @aryan43 for C++ code. Hope to see your knight badge soon.\n\n### Approach\n\n1. **Count the total number of 1s** in the array. This count will be the size of the window we need to consider.\n2. **Form an extended array**: To handle the circular nature, concatenate the array to itself. This allows us to simulate the circular behavior by considering subarrays that wrap around the end to the beginning.\n3. **Sliding Window Technique**: Use a sliding window of size equal to the count of 1s to find the subarray with the minimum number of 0s. This gives the minimum swaps needed.\n\n### Detailed Steps\n\n1. **Initial Setup**:\n   - Count the total number of 1s in the array.\n   - If there are no 1s, no swaps are needed.\n\n2. **Extended Array**:\n   - Concatenate the array with itself to handle the circular nature.\n\n3. **Sliding Window**:\n   - Initialize a window of size equal to the number of 1s.\n   - Count the number of 0s in the initial window.\n   - Slide the window across the extended array and update the count of 0s.\n   - Track the minimum count of 0s in any window of the required size.\n\n4. **Result**:\n   - The minimum count of 0s found in any window is the minimum number of swaps needed.\n\n### Python Solution\n\n```python\nclass Solution(object):\n    def minSwaps(self, nums):\n        """\n        :type nums: List[int]\n        :rtype: int\n        """\n        n = len(nums)\n        count_ones = nums.count(1)\n        if count_ones == 0:\n            return 0\n\n        # Extend the array to handle the circular nature\n        extended_nums = nums + nums\n\n        # Initialize the number of zeroes in the first window\n        current_zeroes = extended_nums[:count_ones].count(0)\n        min_zeroes = current_zeroes\n\n        # Slide the window\n        for i in range(count_ones, len(extended_nums)):\n            if extended_nums[i] == 0:\n                current_zeroes += 1\n            if extended_nums[i - count_ones] == 0:\n                current_zeroes -= 1\n\n            min_zeroes = min(min_zeroes, current_zeroes)\n\n        return min_zeroes\n\n# Test the solution\nsol = Solution()\nnums = [1, 0, 1, 0, 1, 0, 0, 1]\nprint(sol.minSwaps(nums))  # Expected output should consider the circular nature\n```\n\n### C++ Solution\n\n```cpp\nclass Solution {\npublic:\n    int minSwaps(vector<int>& nums) {\n        int n = nums.size();\n        int count_ones = count(nums.begin(), nums.end(), 1);\n        if (count_ones == 0) return 0;\n\n        // Extend the array to handle the circular nature\n        vector<int> extended_nums(nums.begin(), nums.end());\n        extended_nums.insert(extended_nums.end(), nums.begin(), nums.end());\n\n        // Initialize the number of zeroes in the first window\n        int current_zeroes = count(extended_nums.begin(), extended_nums.begin() + count_ones, 0);\n        int min_zeroes = current_zeroes;\n\n        // Slide the window\n        for (int i = count_ones; i < extended_nums.size(); ++i) {\n            if (extended_nums[i] == 0) current_zeroes++;\n            if (extended_nums[i - count_ones] == 0) current_zeroes--;\n\n            min_zeroes = min(min_zeroes, current_zeroes);\n        }\n\n        return min_zeroes;\n    }\n};\n\n// Test the solution\nint main() {\n    Solution sol;\n    vector<int> nums = {1, 0, 1, 0, 1, 0, 0, 1};\n    cout << sol.minSwaps(nums) << endl;  // Expected output should consider the circular nature\n    return 0;\n}\n```\n\n### Java Solution\n\n```java\nclass Solution {\n    public int minSwaps(int[] nums) {\n        int n = nums.length;\n        int countOnes = 0;\n        for (int num : nums) {\n            if (num == 1) countOnes++;\n        }\n        if (countOnes == 0) return 0;\n\n        // Extend the array to handle the circular nature\n        int[] extendedNums = new int[2 * n];\n        System.arraycopy(nums, 0, extendedNums, 0, n);\n        System.arraycopy(nums, 0, extendedNums, n, n);\n\n        // Initialize the number of zeroes in the first window\n        int currentZeroes = 0;\n        for (int i = 0; i < countOnes; i++) {\n            if (extendedNums[i] == 0) currentZeroes++;\n        }\n        int minZeroes = currentZeroes;\n\n        // Slide the window\n        for (int i = countOnes; i < extendedNums.length; i++) {\n            if (extendedNums[i] == 0) currentZeroes++;\n            if (extendedNums[i - countOnes] == 0) currentZeroes--;\n\n            minZeroes = Math.min(minZeroes, currentZeroes);\n        }\n\n        return minZeroes;\n    }\n\n    // Test the solution\n    public static void main(String[] args) {\n        Solution sol = new Solution();\n        int[] nums = {1, 0, 1, 0, 1, 0, 0, 1};\n        System.out.println(sol.minSwaps(nums));  // Expected output should consider the circular nature\n    }\n}\n```\n\n### Explanation\n\n1. **Count the number of 1s**: The initial step counts how many `1`s are in the array.\n2. **Extended Array**: We duplicate the array to handle its circular nature.\n3. **Sliding Window Initialization**: The window is initialized to the size of the number of `1`s, and the count of `0`s in the window is calculated.\n4. **Sliding the Window**: As the window slides across the extended array, the count of `0`s is updated dynamically by adding the new element and removing the old one.\n5. **Tracking Minimum Zeroes**: The minimum count of `0`s found in any window is the result, as it represents the fewest swaps needed to group all `1`s together.\n\n**Time Complexity**: O(n), where n is the number of elements in the original array. This is because we are iterating over the array multiple times in linear time.\n**Space Complexity**: O(n), since we are using an extended array that is twice the size of the original array.
['Array', 'Sliding Window', 'Python', 'C++', 'Java']