C# Solution for Minimum Swaps to Group All 1's Together II Problem
Intuition\n Describe your first thoughts on how to solve this problem. \n1.\tCalculate Total Ones: Count the total number of 1\u2019s (totalOnes) in the array.\
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1.\tCalculate Total Ones: Count the total number of 1\u2019s (totalOnes) in the array.\n2.\tPrefix Sums: Use a prefix sum array to quickly calculate the number of 1\u2019s in any subarray. This helps in efficiently computing the number of 1\u2019s in any sliding window.\n3.\tFind Minimum Swaps: Using the prefix sum array, find the minimum number of swaps required to group all 1\u2019s together by checking all possible windows of size totalOnes.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tCalculate Total Ones:\n\t\u2022\tLoop through the array to count the number of 1\u2019s (totalOnes).\n2.\tPrefix Sum Array:\n\t\u2022\tCreate a prefix sum array where prefixSum[i] contains the sum of elements from the start of the array to the i-1 index.\n\t\u2022\tThis allows quick calculation of the number of 1\u2019s in any subarray by taking the difference between two prefix sums.\n3.\tFind Minimum Swaps:\n\t\u2022\tLoop through the array and check all possible windows of size totalOnes to find the one with the maximum number of 1\u2019s.\n\t\u2022\tUse the prefix sum array to calculate the number of 1\u2019s in each window efficiently.\n\t\u2022\tThe minimum swaps required will be totalOnes - maxOnesInWindow.\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tCalculating Total Ones: This takes O(n).\n\u2022\tPrefix Sum Array Calculation: This takes O(n).\n\u2022\tSliding Window Check: This takes O(n).\nThus, the total time complexity is O(n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tPrefix Sum Array: This requires O(n) space.\n\u2022\tAdditional Variables: A few extra variables are used, which is O(1).\nTherefore, the space complexity is O(n).\n\n# Code\n```\npublic class Solution {\n    public int MinSwaps(int[] nums) {\n        int n = nums.Length;\n        \n        // Step 1: Calculate total number of 1\'s\n        int totalOnes = 0;\n        foreach (int num in nums) {\n            if (num == 1) {\n                totalOnes++;\n            }\n        }\n        \n        // Edge case: if there are no 1\'s or only one 1, no swaps needed\n        if (totalOnes <= 1) {\n            return 0;\n        }\n        \n        // Step 2: Compute the prefix sum array\n        int[] prefixSum = new int[n + 1];\n        for (int i = 0; i < n; i++) {\n            prefixSum[i + 1] = prefixSum[i] + nums[i];\n        }\n        \n        // Step 3: Find the minimum number of swaps required\n        int minSwaps = int.MaxValue;\n        \n        // Check all possible windows of size totalOnes\n        for (int i = 0; i < n; i++) {\n            int end = i + totalOnes - 1;\n            int currentOnes;\n            \n            if (end < n) {\n                currentOnes = prefixSum[end + 1] - prefixSum[i];\n            } else {\n                currentOnes = (prefixSum[n] - prefixSum[i]) + (prefixSum[end - n + 1]);\n            }\n            \n            minSwaps = Math.Min(minSwaps, totalOnes - currentOnes);\n        }\n        \n        return minSwaps;\n    }\n}\n```
['C#']