Answer:
The running time is quadratic (O(n²) )
Step-by-step explanation:
For the set up, we have a constant running time of C. The, a log-linearsorting is called, thus, its execution time, denoted by T(n), is O(n*log(n)). Then, we call n times a linear iteration, with a running time of an+b, for certain constants a and b, thus, the running time of the algorithm is
C + T(n) + n*(a*n+b) = an²+bn + T + C
Since T(n) is O(n*log(n)) and n² is asymptotically bigger than n*log(n), then the running time of the algorith is quadratic, therefore, it is O(n²).