Question

Insertion Sort

Which are true of Insertion Sort (traditional implementation, without optimizations)?

Multiple answers:You can select more than one option.

Please, include the explanation with the answer.

A) It uses Θ(n^2) comparisons in the worst case

B) It uses Θ(n^2) comparisons in the average case

C) It uses Θ(n^2) comparisons in the best case

D) It uses Θ(n^2) movements of elements in the worst case

E) It uses Θ(n^2) movements of elements in the average case

F) It uses Θ(n^2) movements of elements in the best case

Answer 1

Two correct options should be: A and C

Reason: Counting comparisons or swaps(movement of elements) yields similar results. Each time through the inner for loop yields both a comparison and a swap, except the last (i.e., the comparison that fails the inner for loop's test), which has no swap. Thus, the number of swaps for the entire sort operation is n−1 less than the number of comparisons. This is 0 in the best case, and O(n^2) in the average and worst cases.

For reference Insertion sort algorithm goes like:

INSERTION-SORT(A)
   for i = 1 to n                             //outer loop run n times in best case
        key = A [i]
        j = i – 1
         while j > = 0 and A[j] > key             //inner loop for comparison & swapping
                A[j+1] = A[j]
                j = j – 1
        End while 
        A[j+1] = key
  End for 

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