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.kt
fun main() {
val arr = intArrayOf(5, 1, 4, 2, 8)
val n = arr.size
for (i in 0 until n - 1) {
var swapped = false
for (j in 0 until n - i - 1) {
if (arr[j] > arr[j + 1]) {
val tmp = arr[j]
arr[j] = arr[j + 1]
arr[j + 1] = tmp
swapped = true
}
}
if (!swapped) {
break
}
}
println(arr.joinToString(prefix = "[", postfix = "]"))
}
Complexity
- Time: O(n^2) worst and average; O(n) best (already sorted with early exit)
- Space: O(1)
- Stable: yes
Implementation notes
- Kotlin: nested
forloops with the early-exitswappedflag.arr.sort()would hide the comparison-and-swap the lesson is teaching. - The explicit
val tmp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = tmpthree-line swap keeps the move visible without leaning on the(arr[j], arr[j+1]) = (arr[j+1], arr[j])destructuring style. - The replay distinguishes compare frames from swap frames so the
moving pivot value is visible. The pass number and
swappedflag 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.