Selection Sort

time O(n^2)space O(1)

The array splits into a sorted prefix (empty at first) and an unsorted suffix. Each pass finds the SMALLEST value anywhere in the unsorted suffix and swaps it into the very next open slot — so the sorted prefix only ever grows by one confirmed-correct element per pass, and after n-1 passes there's nowhere left for anything to be out of place.

Details

Animated on: Sort nums = [64, 25, 12, 22, 11] in ascending order using selection sort.

Problem:Selection SortEasy

Sorting

step 1 / 9

[0]

[1]

[2]

[3]

[4]

Pass i=0: scan the whole array for its minimum. Start with minIdx=0 (64) as the candidate.

i=0· arr[0] = 64

Pseudocode

1FUNCTION selectionSort(arr):
2  n = the length of arr
3  FOR i from 0 to n-2:
4    minIdx = i   (assume the smallest sits here)
5    FOR j from i+1 to n-1:
6      IF arr[j] < arr[minIdx]:
7        minIdx = j
8    swap arr[i] and arr[minIdx]

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Every step in words

Show all 9 steps

Pass i=0: scan the whole array for its minimum. Start with minIdx=0 (64) as the candidate.
Scanning j=1..4: 25<64→minIdx=1; 12<25→minIdx=2; 22<12? no; 11<12→minIdx=4. Smallest found: 11 at index 4.
Swap arr[0] and arr[4]: [11, 25, 12, 22, 64]. Index 0 is now correctly placed.
Pass i=1: scan indices 1..4. 12<25→minIdx=2; 22<12? no; 64<12? no. Smallest of the suffix: 12 at index 2.
Swap arr[1] and arr[2]: [11, 12, 25, 22, 64]. Index 1 is now correctly placed.
Pass i=2: scan indices 2..4. 22<25→minIdx=3; 64<22? no. Smallest of the suffix: 22 at index 3.
Swap arr[2] and arr[3]: [11, 12, 22, 25, 64]. Index 2 is now correctly placed.
Pass i=3: scan index 4 only. 64<25? no. minIdx stays 3 — arr[3] was already the smallest remaining. Swap with itself (no change).
All 4 passes done (n-1 = 4 for n=5). Sorted: [11, 12, 22, 25, 64].

Now solve it yourself

Watching is the first half. The second is writing it without the animation.

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