**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[0] . . . 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;
/* Head ends here */
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
```

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