Sorting

Selection Sort

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] sorts in two real swaps; the last two passes find minIdx == i and skip the swap.

running minimum Track the index of the smallest value seen during a scan.

Visual walkthrough

The pinned input [5, 1, 4, 2, 8] sorts with two real swaps. The frames keep the running minimum and swap positions visible.

Step 1 - First scan finds 1

In the first pass, min_idx moves from 5 to 1.

First pass over [5, 1, 4, 2, 8]: 1 is the running minimum.i0i1i2i3i451428imin

Step 2 - Swap 5 and 1

The smallest value moves into the first sorted slot.

After swap: [1, 5, 4, 2, 8].i0i1i2i3i415428sorted

Step 3 - Second scan finds 2

In the unsorted suffix, 2 is smaller than 5 and becomes the next minimum.

Second pass: 2 is selected from the suffix.i0i1i2i3i415428sortedimin

Step 4 - Sorted after two swaps

Swapping 5 and 2 gives [1, 2, 4, 5, 8]; later passes find no real swap.

After the second real swap: [1, 2, 4, 5, 8].i0i1i2i3i412458sortedsorted

Basic Implementation

basic.dart 
void main() {
  final arr = <int>[5, 1, 4, 2, 8];
  final n = arr.length;
  for (var i = 0; i < n - 1; i++) {
    var minIdx = i;
    for (var j = i + 1; j < n; j++) {
      if (arr[j] < arr[minIdx]) {
        minIdx = j;
      }
    }
    if (minIdx != i) {
      final 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
  • Swaps: at most n-1

Implementation notes

  • Dart: skip the swap when minIdx == i to match the lesson spec's frame counts. Do not delegate to arr.sort().
  • The replay highlights j (scanning) versus minIdx (running minimum) distinctly, then animates the per-pass swap.