Detailed explained C++ prefix sum Solution.... ✔️✔️✔️
Intuition\n###### Prefix Sum and hashing interval.\n\n# Approach\n\nFunction Purpose: \nThe bestRotation function finds the optimal rotation index to maximize t
# Intuition\n###### Prefix Sum and hashing interval.\n\n# Approach\n\n**Function Purpose:** \nThe bestRotation function finds the optimal rotation index to maximize the number of elements greater than or equal to their indices in a given array a.\n\n**Algorithm:**\n\n1. Initialize variables and a vector ans to store incremental changes in counts.\n2. Iterate through each element of the array a and update the ans vector based on the comparison of array values and their indices.\n3. Calculate the prefix sum of the ans vector.\n4. Find the index (k) with the maximum value in the prefix sum array.\n5. Return the optimal rotation index k.\n\n**Incremental Update:**\n1. If a[i] is less than or equal to i, increment counts in the ans vector for specific indices to reflect the rotation.\n2. If a[i] is greater than i, update counts accordingly.\n\n**Prefix Sum Calculation:**\nAccumulate the changes stored in the ans vector to obtain the prefix sum.\n\n**Finding Maximum:**\nIterate through the prefix sum array to find the index k with the maximum value.\nReturn: The function returns the optimal rotation index k.\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nclass Solution {\npublic:\n    int bestRotation(vector<int>& a) {\n        int n = a.size();\n        vector<int> ans(n+1);\n        for(int i = 0; i<n; ++i)\n        {\n            if(a[i]<=i)\n            ans[0]++, ans[i+1]++, ans[n]--, ans[i-a[i]+1]--;\n            else ans[i+1]++, ans[i+n-a[i]+1]--;\n        }\n        int k = 0;\n        for(int i = 1; i<=n; ++i)\n        {\n            ans[i] += ans[i-1];\n            if(ans[i] > ans[k]) k = i;\n        }\n        return k;\n    }\n};\n```\n\nLike if you find helpful \uD83D\uDC4D\uD83D\uDC4D\uD83D\uDC4D
['Array', 'Prefix Sum', 'C++']