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- Interview Preparation Kit
- Sorting
- Sorting: Bubble Sort

# Sorting: Bubble Sort

# Sorting: Bubble Sort

Consider the following version of Bubble Sort:

```
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
// Swap adjacent elements if they are in decreasing order
if (a[j] > a[j + 1]) {
swap(a[j], a[j + 1]);
}
}
}
```

Given an array of integers, sort the array in ascending order using the *Bubble Sort* algorithm above. Once sorted, print the following three lines:

`Array is sorted in numSwaps swaps.`

, where is the number of swaps that took place.`First Element: firstElement`

, where is the*first*element in the sorted array.`Last Element: lastElement`

, where is the*last*element in the sorted array.

**Hint:** To complete this challenge, you must add a variable that keeps a running tally of *all* swaps that occur during execution.

**Example**

```
swap a
0 [6,4,1]
1 [4,6,1]
2 [4,1,6]
3 [1,4,6]
```

The steps of the bubble sort are shown above. It took swaps to sort the array. Output is:

```
Array is sorted in 3 swaps.
First Element: 1
Last Element: 6
```

**Function Description**

Complete the function *countSwaps* in the editor below.

countSwaps has the following parameter(s):

*int a[n]:*an array of integers to sort

**Prints**

Print the three lines required, then return. No return value is expected.

**Input Format**

The first line contains an integer, , the size of the array .

The second line contains space-separated integers .

**Constraints**

**Output Format**

**Sample Input 0**

```
STDIN Function
----- --------
3 a[] size n = 3
1 2 3 a = [1, 2, 3]
```

**Sample Output 0**

```
Array is sorted in 0 swaps.
First Element: 1
Last Element: 3
```

**Explanation 0**

The array is already sorted, so swaps take place.

**Sample Input 1**

```
3
3 2 1
```

**Sample Output 1**

```
Array is sorted in 3 swaps.
First Element: 1
Last Element: 3
```

**Explanation 1**

The array is *not sorted*, and its initial values are: . The following swaps take place:

At this point the array is sorted and the three lines of output are printed to stdout.