Simple O(N) logic and implementation
Intuition\n Describe your first thoughts on how to solve this problem. \nWe can observe that selecting a position with a 1 and then obtaining neighboring 1s inc
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can observe that selecting a position with a 1 and then obtaining neighboring 1s incurs a cost equal to the distance from that position to the neighboring 1. However, using operation type 1 to increase its value by 1 costs only 2 operations. Neighboring 1s farther than a distance of 2 require more operations than using operation type 1. Therefore, we can efficiently utilize operation type 1, which can be used at most mx times.\n\nTo consider an edge case, we cannot use operation type 1 for neighbors that are just 1 distance away, i.e., adjacent to the current index, because it costs only 1 operation, which is less than the 2 operations of operation type 1. Thus, we consider four cases:\n\nWhen all mx operations are used.\nWhen mx-1 operations are used.\nWhen mx-2 operations are used.\nWhen mx-3 operations are used.\nAll other cases will fully use mx operations to minimize costs.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo solve this, I call a function that determines the remaining 1s which need to be included by corresponding swaps, i.e., the distance of 1s from that position. To solve this efficiently, I use a sliding window containing k ones that are needed, and I slide this window having k ones by maintaining l1 (start), m1 (mid), and r1 (end) of the window. I then slide this window using another concept called "next vector," where I store the index of the next 1s so that I can slide easily.\n\nAfter analyzing the cost while sliding, we can see that when m1 shifts to m2, our costs are reduced by (m1-l1) and increased by (r2-m2). In even k\'ones, I am putting m1 on k/2\'s, so (m2-m1) also decreases.(here r2-new window end,m2-new window mid,l2-new window start).\n\n# Complexity\n- Time complexity:O(N) \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n\n# Code\n```\nclass Solution {\npublic:\n    long long f(vector<int>& nums, int k) {\n        if (k == 0)\n            return 0LL;\n\n        int n = nums.size();\n        int l1 = -1, m1 = -1, r1 = -1;\n        int ctt = 0;\n        int g = k / 2;\n        if (k % 2 != 0)\n            g++;\n        for (int i = 0; i < n; i++) {\n\n            if (nums[i] == 1)\n                ctt++;\n\n            if (ctt == 1 && nums[i] == 1) {\n                l1 = i;\n            }\n\n            if (ctt == g && nums[i] == 1) {\n                m1 = i;\n            }\n\n            if (ctt == k && nums[i] == 1) {\n                r1 = i;\n                break;\n            }\n        }\n\n        vector<int> next(n, -1);\n        int prev = -1;\n\n        for (int i = n - 1; i >= 0; i--) {\n\n            if (nums[i] == 1) {\n                next[i] = prev;\n                prev = i;\n            }\n        }\n        if (r1 == -1)\n            return 1e14;\n\n        long long cost = 0, ct = 0;\n\n        for (int i = 0; i < n; i++) {\n            if (ct == k)\n                break;\n            if (nums[i] == 1)\n                ct++;\n            if (nums[i] == 1 && ct != g) {\n                cost += abs(m1 - i);\n            }\n        }\n        long long ans = 1e14;\n        ans = min(ans, cost);\n\n        while (next[r1] != -1) {\n\n            int m2 = next[m1];\n            int r2 = next[r1];\n\n            int l2 = next[l1];\n\n            long long new_cost = cost - (m1 - l1) + (r2 - m2);\n            if (k % 2 == 0)\n                new_cost -= (m2 - m1);\n\n            ans = min(ans, new_cost);\n            cost = new_cost;\n            r1 = r2;\n            l1 = l2;\n            m1 = m2;\n        }\n\n        return ans;\n    }\n    long long minimumMoves(vector<int>& nums, int k, int mx) {\n\n        int n = nums.size();\n        long long ans = 1e14;\n\n        int case1 = max(0, k - mx);\n        long long c1 = min(mx, k) * 2;\n        ans = min(ans, c1 + f(nums, case1));\n\n        if (mx >= 1) {\n\n            int case2 = max(0, case1 + 1);\n            long long c2 = c1 - 2 + f(nums, case2);\n            ans = min(ans, c2);\n        }\n        if (mx >= 2 && k >= 2) {\n            int case3 = max(0, case1 + 2);\n            long long c3 = c1 - 4 + f(nums, case3);\n            ans = min(ans, c3);\n        }\n        if (mx >= 3 && k >= 3) {\n            int case4 = max(0, case1 + 3);\n            long long c4 = c1 - 6 + f(nums, case4);\n            ans = min(ans, c4);\n        }\n        return ans;\n    }\n};\n```
['C++']