### Problem Statement :

```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:

1. Array is sorted in numSwaps swaps., where numSwap  is the number of swaps that took place.
2. First Element: firstElement, where firstElement  is the first element in the sorted array.
3. Last Element: lastElement, where lastElement  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
a = [ 6, 1, 4 ]

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,n , the size of the array a.
The second line contains n space-separated integers a[ i ].

Constraints

2  <=  n   <=  600
1  <=   a[ i ]  <=  2* 10^6```

### Solution :

```                            ```Solution in C :

In C :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int n;
scanf("%d",&n);
int *a = malloc(sizeof(int) * n);
for(int a_i = 0; a_i < n; a_i++){
scanf("%d",&a[a_i]);
}
int t=0;
for (int i = 0; i < n; i++) {
// Track number of elements swapped during a single array traversal
int numberOfSwaps = 0;

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

// If no elements were swapped during a traversal, array is sorted
if (numberOfSwaps == 0) {
break;
}
}
printf("Array is sorted in %d swaps.\n",t);
printf("First Element: %d\n",a);
printf("Last Element: %d\n",a[n-1]);
return 0;
}```
```

```                        ```Solution in C++ :

In C ++ :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

int main(){
int n,temp,c=0;
cin >> n;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
for(int i=0;i<n-1;i++)
{
for(int j=0;j<n-i-1;j++)
{
if(a[j]>a[j+1])
{
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
c++;
}
}

if(c==0)
{
break;
}}
cout<<"Array is sorted in "<<c<<" swaps."<<endl;
cout<<"First Element:"<<" "<<a<<endl;
cout<<"Last Element:"<<" "<<a[n-1]<<endl;
return 0;
}```
```

```                        ```Solution in Java :

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int a[] = new int[n];
for(int a_i=0; a_i < n; a_i++){
a[a_i] = in.nextInt();
}
bubbleSort(a);
}

public static void bubbleSort(int[] a){
int numSwaps = 0;
for(int i=0; i< a.length; i++){
for(int j=0; j < a.length - 1; j++){
if(a[j] > a[j+1]){
int tmp = a[j];
a[j] = a[j+1];
a[j+1] = tmp;
numSwaps++;
}

}
if(numSwaps==0)
break;
}
System.out.println("Array is sorted in " + numSwaps +" swaps.");
System.out.println("First Element: " + a);
System.out.println("Last Element: " + a[a.length-1]);

}
}```
```

```                        ```Solution in Python :

In Python3 :

n = int(input().strip())
a = list(map(int, input().strip().split(' ')))
swap=0
for i in range (0,len(a)):
for j in range (0,len(a)-1):
if a[j]>a[j+1]:
temp=a[j]
a[j]=a[j+1]
a[j+1]=temp
swap+=1
print("Array is sorted in",swap,"swaps.")
print("First Element:",a)
print("Last Element:",a[len(a)-1])```
```

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing