Final answer:
In the QuickSort algorithm with the first element as pivot, in the given arrays, the number of comparisons t1 for {2, 3, 5, 1, 4} is 7 and t2 for {4, 1, 5, 3, 2} is 6. Comparing both, we find that t1 > t2, which answers the question.
Step-by-step explanation:
The question is related to the QuickSort algorithm used in the sorting of an array when the first element is chosen as the pivot. To determine the number of comparisons t1 for the input {2, 3, 5, 1, 4}, we must perform the algorithm step by step. The first comparison will involve comparing each of the other elements to the pivot (2), resulting in four comparisons. Now, since '1' is less than '2', we will swap them and '2' becomes the pivot for the subset {1}. Since there are no more elements to compare with in this subset, no further comparisons are made here. For the subset {3, 5, 4} with '3' as the pivot, two comparisons are made. Finally, the subset {5, 4} involves one more comparison. Thus, t1 = 4 + 2 + 1 = 7 comparisons in total.
The number of comparisons t2 for the input {4, 1, 5, 3, 2} is calculated in a similar fashion. The pivot '4' will be compared with four elements, resulting in 4 comparisons. The subset {1, 3, 2} will have '1' as the pivot, but since '1' is the smallest, it makes no comparisons, and the subset {3, 2} will make 1 comparison. Then '3' will have '2' to compare with, resulting in one more comparison. Therefore, t2 = 4 + 1 + 1 = 6 comparisons.
Comparing t1 and t2 shows that t1 > t2. Hence the answer to the question is C. t1 > t2.