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.java 
public class Basic {
    public static void main(String[] args) {
        int[] arr = {5, 1, 4, 2, 8};
        int n = arr.length;
        for (int i = 0; i < n - 1; i++) {
            boolean swapped = false;
            for (int j = 0; j < n - i - 1; j++) {
                if (arr[j] > arr[j + 1]) {
                    int tmp = arr[j];
                    arr[j] = arr[j + 1];
                    arr[j + 1] = tmp;
                    swapped = true;
                }
            }
            if (!swapped) {
                break;
            }
        }
        System.out.println(java.util.Arrays.toString(arr));
    }
}

Complexity

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

Implementation notes

  • Java: nested for loops with the early-exit swapped flag. Never call Arrays.sort(...); the lesson is teaching the comparison-and-swap.
  • 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.