比较选择排序中的平均情况和最坏情况交换的数量

Question

I came across following problem:

State true or false:
In selection sort, total swaps in average case >= total swaps in worst case

After pondering a bit and trying out some examples, I reached following conclusions:

In selection sort, worst case = best case = avg case = O(n^2).

In selection sort, time complexity is determined by number of comparisons which is always n(n-1)/2 = O(n^2)

So we cannot say, total swaps in average case>=total swaps in worst case. On the other hand, minimum number of swaps happen when array is already sorted. And maximum number of swaps happen when first half of the array contains n/2 largest elements in reverse sorted order and second half of the array contains n/2 smallest elements in sorted order. However this will not necessarily always lead to n-1 swaps. Instead it might lead to n-1, n-2 or even n-3 swaps as explained in the figure below:

I am correct with above thinking?

Answer 1

An average case cannot exceed the worst case, this is impossible.

The average case can equal the worst case, provided all cases have the same complexity.

Depending on how exactly you implement SelectionSort, you have two situations:

• if you "blindly" swap after every selection, the count of swaps is N-1 and is constant;

• if you swap only when the selected element is not already in place, the count of swaps can be less (in the best case - a sorted array - the count is 0).