# Insertion Sort - Part 2

### Problem Statement :

```In Insertion Sort Part 1, you inserted one element into an array at its correct sorted position. Using the same approach repeatedly, can you sort an entire array?

Guideline: You already can place an element into a sorted array. How can you use that code to build up a sorted array, one element at a time? Note that in the first step, when you consider an array with just the first element, it is already sorted since there's nothing to compare it to.

In this challenge, print the array after each iteration of the insertion sort, i.e., whenever the next element has been inserted at its correct position. Since the array composed of just the first element is already sorted, begin printing after placing the second element.

Function Description

Complete the insertionSort2 function in the editor below.

insertionSort2 has the following parameter(s):

int n: the length of arr
int arr[n]: an array of integers
Prints

At each iteration, print the array as space-separated integers on its own line.

Input Format

The first line contains the integer n, the size of the array arr.
The next line contains n space-separated integers arr  . . . arr[ n - 1 ] .

Constraints

1  <=  n  <=  1000
-10000 <=  arr[ i ]  <=  10000

Output Format

Print the entire array on a new line at every iteration.```

### Solution :

```                            ```Solution in C :

In   C++  :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;

void insertionSort(vector <int>  ar) {
int n = ar.size();
for(int i=1;i<n;i++){
int curr = ar[i];
for(int j=i-1;j>=0;j--){
if(ar[j]>curr){
ar[j+1]=ar[j];
if(j==0)
ar[j]=curr;
}
else{
ar[j+1]=curr;
j=-1;
}
}
for(int t=0;t<n;t++)
cout<<ar[t];
cout<<endl;
}
}

/* Tail starts here */
int main() {
vector <int>  _ar;
int _ar_size;
cin >> _ar_size;
for(int _ar_i=0; _ar_i<_ar_size; _ar_i++) {
int _ar_tmp;
cin >> _ar_tmp;
_ar.push_back(_ar_tmp);
}
insertionSort(_ar);
return 0;
}

In   Java  :

import java.util.Scanner;

public class Solution {

public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
int s=scan.nextInt();
int ar[]=new int[s];
boolean check=false;
for(int i=0;i<s;i++)
{
ar[i]=scan.nextInt();
}
for(int i=1;i<s;i++)
{
int temp=ar[i];
for(int j=i-1;j>=0 && temp<ar[j];j--)
{
ar[j+1]=ar[j];
ar[j]=temp;
}
for(int j=0;j<s;j++)
System.out.print(ar[j]+" ");
System.out.println();
}
}
}

In   C  :

#include<stdio.h>
int main()
{
int size,num,i,j,t;
int *x;
scanf("%d",&size);
x=(int *)malloc(sizeof(int)*size);
for(i=0;i<size;i++)
scanf("%d",&x[i]);
//insertion sort..
for(i=1;i<size;i++)
{
num=x[i];
for(j=i-1;j>=0&&num<x[j];j--)
x[j+1]=x[j];
x[j+1]=num;
for(t=0;t<size;t++)
printf("%d ",x[t]);
printf("\n");
}

}

In   Python3  :

size = int(input())
array = input().split(" ")
arr = ['None'] * size
for i in range(size):
arr[i] = int(array[i])
i = 1
while i < size:
tmp = arr[i]
j = i - 1
while arr[j] > tmp and j > -1:
arr[j+1] = arr[j]
j = j - 1
arr[j+1] = tmp
for k in range(size):
print(arr[k],end = " ")
print()
i = i + 1```
```

## 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

## Tree: Postorder Traversal

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

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.