Minimum Operations Solution without creating array | without building array

Solving the Mininum Operations Solution in O(1) memory
'''

    int mid = n/2;
    int count =0;
   
    if(n%2 ==0){
        int number = (((2*(mid-1))+1) + ((2*mid)+1))/2;
        for(int i =0 ;i< n/2;i++){
            int temp = (2*i)+1;
            count += Math.Abs(number - temp);
        }
    }
    else{
        int number = (2*mid) +1;
        for(int i=0;i<n/2;i++){
            int temp = (2*i)+1;
            count += Math.Abs(number - temp);
        }
    }
    
    return count;

'''

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