Repeatedly find the index of the smallest remaining element and swap it into the next "sorted prefix" slot. Unlike bubble sort, only one swap per pass.

Algorithm

Canonical input [5, 1, 4, 2, 8] finishes after four passes, with two real swaps (passes 0 and 1) and two skip-swap passes (minIdx == i). Final array [1, 2, 4, 5, 8].

Basic Implementation

basic.kt
fun main() {
	val arr = intArrayOf(5, 1, 4, 2, 8)
	val n = arr.size
	for (i in 0 until n - 1) {
		var minIdx = i
		for (j in i + 1 until n) {
			if (arr[j] < arr[minIdx]) {
				minIdx = j
			}
		}
		if (minIdx != i) {
			val tmp = arr[i]
			arr[i] = arr[minIdx]
			arr[minIdx] = tmp
		}
	}
	println(arr.joinToString(prefix = "[", postfix = "]"))
}

Complexity

  • Time: O(n^2) regardless of input order
  • Space: O(1)
  • Stable: no
  • Swap count: at most n-1

Implementation notes

  • Kotlin: same loop shape as Python / Java / JavaScript / C++ / C / Go / Rust / C#. The if (minIdx != i) guard is the canonical skip-swap variant from the lesson spec.
  • var minIdx = i keeps the running-minimum invariant visible; the three-line val tmp = arr[i]; arr[i] = arr[minIdx]; arr[minIdx] = tmp swap mirrors the bubble-sort lesson.
  • The replay highlights the current minIdx distinctly from the scanning index j so the viewer sees the running minimum travel.
running minimum `minIdx` tracks the index of the smallest value seen in `arr[i..]`.
sorted prefix After each pass, `arr[0..i]` is the final sorted prefix.