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.swift
var arr = [5, 1, 4, 2, 8]
let n = arr.count
for i in 0..<(n - 1) {
	var minIdx = i
	for j in (i + 1)..<n {
		if arr[j] < arr[minIdx] {
			minIdx = j
		}
	}
	if minIdx != i {
		let tmp = arr[i]
		arr[i] = arr[minIdx]
		arr[minIdx] = tmp
	}
}
print(arr)

Complexity

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

Implementation notes

  • Swift: same loop shape as Python / Java / JavaScript / C++ / C / Go / Rust / C# / Kotlin. 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 let 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.