减少两个嵌套for循环的时间复杂度O(N*N)

Question

I am trying to reduce time complexity of the following nested loop which currently has an O(N*N) time complexity:

        for(i = 0; i < N-1; i++){
            for(j = i+1; j < N; j++){
                if((A[j] > B[i])){
                    ctr++; //counting elements satisfying the condition
                }
            }
        }

A and B are just two vectors. I expect reducing O(N*N) to O(N). In addition, I doubt that if sorting A and B will help or not! Thanks!

Answer 1

If you are comparing N*N/2 arbitrary elements, you need N*N/2 operations, ie O(n²). Sorting is O(N*log(N)) and can be done separately on both vectors. Sorting 2 vectors is still O(N*log(N)) because constant factors do not play a role in Big-O notation.

If the vectors are already sorted, you could break the loop after the first element which does not satisfy the condition.