Two Pointers + Prefix Sum
The problem is interesting and the final solution is straightforward, but it involves using two pointers with prefixes. If you\'ve solved this problem before, y
The problem is interesting and the final solution is straightforward, but it involves using two pointers with prefixes. If you\'ve solved this problem before, you might have encountered it. But decoding the given problem statement into a reduced problem is where the complexity lies. That\'s exactly what we\'ll see - how to decode this problem step by step into its solution.\n\nUnderstanding the Problem\n+ To solve this problem, we first determine the maximum possible value of K that can be collected, considering the Max changes and the sum of ones in the original array. \n+ We then look at the minimum moves needed to pick K ones by considering adjacent ones, change operations, and finding the optimal convergence position for the ones in the array. \n+ By iterating over all possible convergence positions, we calculate the minimum number of moves required to collect K ones, taking into account Max changes and adjacent ones.\n\n## So here are key observations:\n\t1) Performing operation-2 on max consecutive one is the 1st optimal way.\n\t2) Then, performing operation-1 is 2nd optimial way. -- i.e If you utilize max changes, you\'re guaranteed that you\'ll get `maxChanges` ones in `maxChnages * 2` moves. -- Hence it is a 2nd optimial way.\n\t3) Then, we should go for remaining ones using operation 2.\n\n### Step-1: Using Max Consecutive Ones and Max Changes\n\tCheck for Max consecutive ones.\n\tApply Max changes if needed.\n\tReturn the minimum moves required.\n\n### Step-2: Utilizing Nearest Neighbors for Remaining Ones\n\tDetermine the ones to collect using adjacent neighbors.\n\tCalculate the prefix sum of positions.\n\tIterate over all possible positions to find the optimal convergence.\n\tAdd the moves required for convergence to the existing moves. \n\tReturn the total minimum moves needed.\n\n### Convergence Function\n\tCalculate the minimum moves needed to converge a set of ones to a specific position.\n\tConsider the positions of ones in the left and right.\n\tSubtract the sum of positions from the product of positions and nu mber of ones.\n\tReturn the total moves required for convergence.\n\n# CODE:\n```\n#define ll long long\nclass Solution \n{\nprivate:  \n    int n;\n    vector<ll> ones;\n    vector<ll> prefix;\n    \npublic:\n    long long minimumMoves(vector<int>& nums, int k, int maxChanges) \n    {\n        n = nums.size();\n        \n        // STEP-1: Identify max consecutive ones.\n        int max_consecutive = 0;\n        for (int i = 0; i < n; i ++) \n        {\n            int cnt = 0;\n            for (int j = 0; j < 3 && i+j < n; j ++) \n            {\n                if ( nums[j+i] == 1 ) cnt ++;\n                else break;\n            }\n            max_consecutive = max (max_consecutive, cnt);\n        }\n                \n        // STEP-2: if maxChanges & consecutive ones satisfies k, then do it.\n        if ( max_consecutive+maxChanges >= k ) \n        {\n            if (k <= max_consecutive) return (k-1);\n            \n            int req_changes = k - max_consecutive;\n            int ans = req_changes*2;\n            if ( max_consecutive > 0 ) ans += (max_consecutive - 1);\n            return ans;\n        }\n\n        // STEP-3: Utilizing Nearest Neighbors for Remaining Ones\n        ones = {-1};\n        prefix = {0};\n        for ( int i = 0; i<n; i++ )\n            if (nums[i] == 1) {\n                ones.push_back(i);\n                prefix.push_back(prefix.back() + i);\n            }\n        \n        int req_1 = k - maxChanges;\n        ll ans = 1e18;\n        \n        for ( int l=1, r=req_1; r<ones.size(); l++, r++ ) \n        {\n            int best_pos = (l + r)/2;\n            ans = min (ans, Converge(l, r, best_pos));\n        }\n        return ans + maxChanges*2;\n    }\n\nprivate:\n    ll Converge (int l, int r, int pos) \n    {        \n        ll left = (pos - l + 1) * ones[pos] - (prefix[pos] - prefix[l-1]);\n        ll right = (prefix[r] - prefix[pos]) - (r - pos) * ones[pos];\n\n        return (left + right);\n    }\n};\n```\n\n\n
['Two Pointers', 'C', 'Prefix Sum']