# Sorting: Bubble Sort

### 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[0]);
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[0]<<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[0]);
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[0])
print("Last Element:",a[len(a)-1])```
```

## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

## Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

## Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

## Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's