Sorting

Bubble Sort

Repeatedly walk the array comparing adjacent pairs and swapping any that are out of order. After pass k, the k largest elements are in their final positions at the end. Stop early when a full pass makes zero swaps.

Algorithm

Canonical input [5, 1, 4, 2, 8] finishes after three passes: two with swaps, then a clean pass that triggers the early exit. 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 swapped = false
	for j in 0..<(n - i - 1) {
		if arr[j] > arr[j + 1] {
			let tmp = arr[j]
			arr[j] = arr[j + 1]
			arr[j + 1] = tmp
			swapped = true
		}
	}
	if !swapped {
		break
	}
}
print(arr)

Complexity

  • Time: O(n^2) worst and average; O(n) best (already sorted with early exit)
  • Space: O(1)
  • Stable: yes

Implementation notes

  • Swift: nested for loops with the early-exit swapped flag. arr.sort() would hide the comparison-and-swap the lesson is teaching.
  • The explicit let tmp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = tmp three-line swap keeps the move visible without leaning on the arr.swapAt(j, j+1) helper.
  • The replay distinguishes compare frames from swap frames so the moving pivot value is visible. The pass number and swapped flag appear in the trace.
adjacent-pair compare and swap Inner loop walks `j` from `0` to `n - i - 2` comparing `arr[j]` and `arr[j + 1]`.
early exit A `swapped` flag set false at the start of each pass. If no swap happened, break out of the outer loop.