Why is Selection Sort said to have O(n) swaps?

Question

I am reading about use cases of Selection Sort, and this source says:

(selection sort is used when...) cost of writing to a memory matters like in flash memory (number of writes/swaps is O(n) as compared to O(n2) of bubble sort)

We can even see O(n^2) swaps in this example: [1, 2, 3, 4, 5]. It's going to have 4 swaps, then 3, then 2, and 1. That is O(n^2), not O(n) swaps. Why do they say the opposite?

Answer 1

A selection sort has a time complexity of O(n²), but only O(n) swaps.

In each iteration i, you go over all the remaining items (in indexes i and onwards), find the right value to populate that index, and swap it there. So in total you perform O(n²) comparisons, but only O(n) swaps.