C++ ||  0ms || 1 line solution || O(1) solution
It is a well known fact that the minimum absolute dist of the numbers in the array is from their median.\n\nWe can observe the fact of the median will be equal 
It is a well known fact that the minimum absolute dist of the numbers in the array is from their median.\n\nWe can observe the fact of the median will be equal to it\'s size i.e median==n/2.\n\nNow we can also observe the array is cyclic about the median meaning that for every indices , there exist another indice in the array having the same absolute difference with respect to the median , so if we calculate for the first n/2 terms we can get the answer.\n\nWe will now having a simple ap where the ans is the sum of the ap .\nThe ans == sum of(n-1 , n-3,n-5,n-7)\n\nIt is a well known fact that the sum of first k (here n/2 )odd numbers is k^2(Proof is left to the reader to prove). This means that our answer is k*(n-k) , where k is the number of terms before the median , =n/2;\n\nhence the answer is simply (n/2)*(n-n/2)\n\n\n```cpp\nclass Solution {\npublic:\n    int minOperations(int n) {\n        return n/2*(n-n/2);\n    }\n};\n\n```
['C++']