Sorting — arranging items in order — is the most fundamental task in computation. Sorting enables efficient searching algorithms such as binary search.

Selection, insertion and bubble sort are easily understandable and also similar to each other, but they are less efficient than merge sort or quick sort. The basic ideas are as below:

**Selection sort**: repeatedly pick the smallest element to append to the result.**Insertion sort**: repeatedly add new element to the sorted result.**Bubble sort**: repeatedly compare neighbor pairs and swap if necessary.

Here we wish to sort an array of integers in Java; this can be easily extended to other data types and collections.

### Selection Sort

*Selection sort* is to repetitively pick up the smallest element and put it into the right position:

- Find the smallest element, and put it to the first position.
- Find the next smallest element, and put it to the second position.
- Repeat until all elements are in the right positions.

A loop through the array finds the smallest element easily. After the smallest element is put in the first position, it is fixed and then we can deal with the rest of the array. The following implementation uses a *nested loop* to repetitively pick up the smallest element and swap it to its final position. The `swap()`

method exchanges two elements in an array.

public static void selectionSort(int[] arr) { // find the smallest element starting from position i for (int i = 0; i < arr.length - 1; i++) { int min = i; // record the position of the smallest for (int j = i + 1; j < arr.length; j++) { // update min when finding a smaller element if (arr[j] < arr[min]) min = j; } // put the smallest element at position i swap(arr, i, min); } } public static void swap (int[] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; }

The following tracks the code on an array with elements `{38, 27, 43, 3, 9, 82, 10}`

.

{38, 27, 43, 3, 9, 82, 10}, i: 0, min: 3, minValue: 3 { 3, 27, 43, 38, 9, 82, 10}, i: 1, min: 4, minValue: 9 { 3, 9, 43, 38, 27, 82, 10}, i: 2, min: 6, minValue: 10 { 3, 9, 10, 38, 27, 82, 43}, i: 3, min: 4, minValue: 27 { 3, 9, 10, 27, 38, 82, 43}, i: 4, min: 4, minValue: 38 { 3, 9, 10, 27, 38, 82, 43}, i: 5, min: 6, minValue: 43 { 3, 9, 10, 27, 38, 43, 82}, i: 6

Suppose the input array has *n* elements. The outer loop runs *n-1* rounds, roughly one for each position. Inside each outer loop, the inner loop goes through the unsorted part of the array. On average, each inner loop scans through *n/2* elements. The total time is roughly *t(n) = (n – 1) * (n/2) *k = O(n^2)*, where *k* denotes the number of basic operations inside each inner loop; the constants are absorbed in the big-Oh notion. Note that in the big-Oh notion, we only care about the dominating factor (which is *n^2* in this case).

### Insertion Sort

*Insertion sort* maintains a sorted sub-array, and repetitively inserts new elements into it. The process is as following:

- Take the first element as a sorted sub-array.
- Insert the second element into the sorted sub-array (shift elements if needed).
- Insert the third element into the sorted sub-array.
- Repeat until all elements are inserted.

The following `insertionSort()`

method implements insertion sort. It uses a nested loop to repetitively insert elements into the sorted sub-array.

public static void insertionSort(int[] arr) { for (int i = 1; i < arr.length; i++) { // a temporary copy of the current element int tmp = arr[i]; int j; // find the position for insertion for (j = i; j > 0; j--) { if (arr[j - 1] < tmp) break; // shift the sorted part to right arr[j] = arr[j - 1]; } // insert the current element arr[j] = tmp; } }

The following tracks the code on an array with elements `{38, 27, 43, 3, 9, 82, 10}`

.

{38, 27, 43, 3, 9, 82, 10} {27, 38, 43, 3, 9, 82, 10}, i: 1 {27, 38, 43, 3, 9, 82, 10}, i: 2 { 3, 27, 38, 43, 9, 82, 10}, i: 3 { 3, 9, 27, 38, 43, 82, 10}, i: 4 { 3, 9, 27, 38, 43, 82, 10}, i: 5 { 3, 9, 10, 27, 38, 43, 82}, i: 6

Suppose the array length is n. The outer loop runs roughly n times, and the inner loop on average runs n/2 times. The total time is about t(n) = n * (n/2) = O(n^2). In terms of the efficiency, this is the same as selection sort.

### Bubble Sort

Bubble sort repetitively compares adjacent pairs of elements and swaps if necessary.

- Scan the array, swapping adjacent pair of elements if they are not in relative order. This bubbles up the largest element to the end.
- Scan the array again, bubbling up the second largest element.
- Repeat until all elements are in order.

The following `bubbleSort()`

method implements bubble sort. It uses a nested loop to repetitively swap elements and bubble up the largest elements one by one.

public static void bubbleSort (int[] data) { for (int i = data.length - 1; i >= 0; i--) { // bubble up for (int j = 0; j <= i - 1; j++) { if (data[j] > data[j + 1]) swap(data, j, j + 1); } } }

The following tracks the code on an array with elements `{38, 27, 43, 3, 9, 82, 10}`

for three rounds of bubbling.

{38, 27, 43, 3, 9, 82, 10} i: 6 {27, 38, 43, 3, 9, 82, 10}, j: 0 {27, 38, 43, 3, 9, 82, 10}, j: 1 {27, 38, 3, 43, 9, 82, 10}, j: 2 {27, 38, 3, 9, 43, 82, 10}, j: 3 {27, 38, 3, 9, 43, 82, 10}, j: 4 {27, 38, 3, 9, 43, 10, 82}, j: 5 i: 5 {27, 38, 3, 9, 43, 10, 82}, j: 0 {27, 3, 38, 9, 43, 10, 82}, j: 1 {27, 3, 9, 38, 43, 10, 82}, j: 2 {27, 3, 9, 38, 43, 10, 82}, j: 3 {27, 3, 9, 38, 10, 43, 82}, j: 4 i: 4 {3, 27, 9, 38, 10, 43, 82}, j: 0 {3, 9, 27, 38, 10, 43, 82}, j: 1 {3, 9, 27, 38, 10, 43, 82}, j: 2 {3, 9, 27, 10, 38, 43, 82}, j: 3

Suppose the array length is n. The outer loop runs roughly n times, and the inner loop on average runs n/2 times. The total time is about *t(n) = n * (n/2) = O(n^2)*. In terms of the efficiency, this is the same as selection sort and insertion sort.

### Related Posts

### References

The following links give visualizations on selection, insertion and bubble sort:

http://www.sorting-algorithms.com/selection-sort

http://www.sorting-algorithms.com/insertion-sort

http://www.sorting-algorithms.com/bubble-sort