Selection Sort places one element into its final position during every iteration.
For an array of $n$ elements:
- Pass $1$ places one element correctly.
- Pass $2$ places another element correctly.
- This continues until only one element remains.
Therefore, only $n-1$ passes are required.
Since the standard Selection Sort implementation performs one swap at the end of each pass:
$\text{Number of swaps}=n-1$
Note : An important feature of Selection Sort is that the number of swaps does not depend on whether the input is sorted, reverse sorted, or randomly ordered. Selection Sort uses only about $N$ exchanges.
