# Hackerrank - Sorting: Bubble Sort Solution

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.

For example, given a worst-case but small array to sort: we go through the following steps:

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

It took swaps to sort the array. Output would beArray is sorted in 3 swaps. First Element: 1 Last Element: 6

**Function Description**

Complete the function *countSwaps* in the editor below. It should print the three lines required, then return.

countSwaps has the following parameter(s):

*a*: an array of integers .

**Input Format**

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

The second line contains space-separated integers .

**Constraints**

**Output Format**

You must print the following three lines of output:

`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.

**Sample Input 0**

```
3
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 and we print the necessary three lines of output shown above.

**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 we print the necessary three lines of output shown above.

### Solution in Python

```
from itertools import product
def countSwaps(a):
swaps = 0
for i,j in product(range(len(a)),range(len(a)-1)):
if a[j]>a[j+1]:
a[j],a[j+1] = a[j+1],a[j]
swaps += 1
print("Array is sorted in %s swaps." %swaps)
print("First Element:", a[0])
print("Last Element:", a[-1])
n = int(input())
a = list(map(int,input().split()))
countSwaps(a)
```

Additional Info

The following two gives same output

```
for i in range(n):
for j in range(n-1):
print(i,j)
```

```
from itertools import product
for i,j in product(range(n),range(n-1)):
print(i,j)
```