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Intuition\n\n Describe your first thoughts on how to solve this problem. \nUnderstanding the Problem: Recognize that we need to rearrange the array such that al
# Intuition\n![Screenshot 2024-08-02 072313.png](https://assets.leetcode.com/users/images/992bb67b-29c5-4aed-b37d-541c27c4e498_1722563645.5521364.png)\n<!-- Describe your first thoughts on how to solve this problem. -->\nUnderstanding the Problem: Recognize that we need to rearrange the array such that all 1s are grouped together, and we want to do this with the minimum number of swaps. The array is circular, meaning the first and last elements are considered adjacent.\n\nCounting 1s: Start by counting the total number of 1s (oneCount) in the array. If oneCount is 0 or 1, no swaps are needed (return 0).\n\nSliding Window Approach: Utilize a sliding window of size oneCount to evaluate each possible segment of the array where all 1s could potentially be grouped together.\n\nCalculating Swaps: For each position of the sliding window:\n\nCalculate the number of 1s in the current window.\nDetermine the number of 0s that would need to be swapped to move all 1s into a contiguous block in this window.\nCircular Consideration: Since the array is circular, after reaching the end of the array, continue counting from the beginning. This requires careful handling to ensure all possible positions are evaluated.\n\nMinimize Swaps: Track the minimum number of swaps required across all valid window positions.\n\nSteps to Implement:\nStep 1: Count the total number of 1s in the array (oneCount).\nStep 2: Handle edge cases where oneCount is 0 or 1.\nStep 3: Initialize a sliding window to calculate the initial number of swaps required for the first segment of size oneCount.\nStep 4: Slide the window across the array, updating the count of 1s and calculating the corresponding swaps for each window position.\nStep 5: Use modulo arithmetic to handle the circular nature of the array.\nStep 6: Return the minimum number of swaps found across all window positions.\nBy following these steps, we can efficiently determine the minimum number of swaps required to group all 1s together in the circular array nums. This approach ensures that we consider all possible ways to rearrange the array while minimizing computational complexity.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n->Count 1s: First, count the total number of 1s (oneCount) in the array. If oneCount is less than 2, return 0 because no swaps are needed.\n\n->Initial Window Sum: Use a sliding window of size oneCount to calculate the sum of 1s in the initial window of the array.\n\n->Calculate Initial Swaps: Compute the number of 0s in the initial window that need to be swapped to group all 1s together initially.\n\nSliding Window Technique:\n\nMove the window across the array to simulate different starting positions where all 1s can be grouped together.\nFor each new position of the window:\nSubtract the contribution of the element that is sliding out of the window.\nAdd the contribution of the element that is sliding into the window.\nCalculate the number of swaps required to group all 1s together for the current window position.\nUpdate the minimum number of swaps required (minSwaps).\nCircular Consideration:\n\nSince the array is circular, after reaching the end of the array, continue counting from the beginning to maintain the circular nature.\nReturn Result: After iterating through all possible positions, minSwaps will contain the minimum number of swaps required to group all 1s together optimally in the circular array nums.\n\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int minSwaps(int[] nums) {\n        int oneCount = 0;\n        int n = nums.length;\n        for (int num : nums) {\n            if (num == 1)oneCount++;\n        }\n        if (oneCount <= 1)\n            return 0;\n        int minSwaps = Integer.MAX_VALUE;\n        int windowSum = 0;\n        for (int i = 0; i < oneCount; i++) {\n            windowSum += nums[i];\n        }\n        minSwaps = oneCount - windowSum;\n        for (int i = 1; i <= n; i++) {\n            windowSum -= nums[i - 1];\n            windowSum += nums[(i + oneCount - 1) % n];\n            int swapsNeeded = oneCount - windowSum;\n            minSwaps = Math.min(minSwaps, swapsNeeded);\n        }\n        \n        return minSwaps;\n    }\n}\n\n```
['Array', 'Sliding Window', 'Java']