C++ | Efficient Minimum Swaps to Group All 1's Together II | 49ms Beats [99.02%]
Intuition\nTo minimize the number of swaps needed to group all 1\'s in a circular binary array, we need to find the optimal position where all 1\'s can be group
# Intuition\nTo minimize the number of swaps needed to group all $$1$$\'s in a circular binary array, we need to find the optimal position where all $$1$$\'s can be grouped together with the fewest swaps. The challenge is to consider the circular nature of the array, which means that the end of the array is adjacent to the beginning.\n\n# Approach\nFirst, we count the total number of $$1$$\'s in the array. Then, we use a sliding window approach to find the maximum number of $$1$$\'s that can be grouped together in any window of length equal to the total count of $$1$$\'s. This is done by extending the array to simulate the circular property. For each possible window, we track the number of $$1$$\'s in the current window and update the maximum number of $$1$$\'s found. Finally, the minimum swaps required are calculated as the difference between the total number of $$1$$\'s and the maximum number of $$1$$\'s that can be grouped together.\n\n# Complexity\n- **Time complexity** $$(O(n))$$, where $$(n)$$ is the length of the array. We traverse the array and its extended version once.\n- **Space complexity** $$(O(1))$$. We use a constant amount of extra space for variables.\n\n\n# Code\n```\nclass Solution {\npublic:\n    int minSwaps(vector<int>& nums) {\n        int ones = count(nums.begin(), nums.end(), 1), maxOnes{0}, winOnes{0}, n = nums.size();\n        if (ones == 0) return 0;\n        for (int i{0}; i < 2 * n; ++i) {\n            winOnes += nums[i % n];\n            if (i >= ones) winOnes -= nums[(i - ones) % n];\n            maxOnes = max(maxOnes, winOnes);\n        }\n        return ones - maxOnes;\n    }\n};\n\n```\n\n## **If you found this solution helpful, please upvote! \uD83D\uDE0A**
['C++']