# Insertion Sort – Part 1 – HackerRank Solution

In this post, we will solve Insertion Sort – Part 1 – HackerRank Solution. This problem (Insertion Sort – Part 1) is a part of HackerRank Problem Solving series.

Sorting
One common task for computers is to sort data. For example, people might want to see all their files on a computer sorted by size. Since sorting is a simple problem with many different possible solutions, it is often used to introduce the study of algorithms.

Insertion Sort
These challenges will cover Insertion Sort, a simple and intuitive sorting algorithm. We will first start with a nearly sorted list.

Insert element into sorted list
Given a sorted list with an unsorted number c in the rightmost cell, can you write some simple code to insert e into the array so that it remains sorted?

Since this is a learning exercise, it won’t be the most efficient way of performing the insertion. It will instead demonstrate the brute-force method in detail.

Assume you are given the array arr = [1, 2, 4, 5, 3] indexed 0 . . . 4. Store the value of arr[4]. Now test lower index values successively from 3 to 0 until you reach a value that is lower than arr[4], at arr[1] in this case. Each time your test fails, copy the value at the lower index to the current index and print your array. When the next lower indexed value is smaller than arr[4], insert the stored value at the current index and print the entire array.

Example

n = 5
arr = [1, 2, 4, 5, 3]
Start at the rightmost index. Store the value of arr[4] = 3. Compare this to each element to the left until a smaller value is reached. Here are the results as described:

``````1 2 4 5 5
1 2 4 4 5
1 2 3 4 5``````

Function Description

Complete the insertionSort1 function in the editor below.

insertionSort1 has the following parameter(s):

• n: an integer, the size of arr
• arr: an array of integers to sort

Returns

• None: Print the interim and final arrays, each on a new line. No return value is expected.

## 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 array as a row of space-separated integers each time there is a shift or insertion.

Sample Input

``````5
2 4 6 8 3``````

Sample Output

``````2 4 6 8 8
2 4 6 6 8
2 4 4 6 8
2 3 4 6 8 ``````

Explanation

3 is removed from the end of the array.
In the 1st line 8 > 3, so 8 is shifted one cell to the right.
In the 2nd line 6 > 3, so 6 is shifted one cell to the right.
In the 3rd line 4 > 3, so 4 is shifted one cell to the right.
In the 4th line 2 < 3, so 3 is placed at position 1.

## Solution – Insertion Sort – Part 1 – HackerRank Solution

### 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;
#define INT_MIN -1000000
void output(const vector<int>& a){
cout << a[1];
for(int j = 2; j < a.size(); ++j){
cout << " " << a[j];
}
cout << endl;
}
void insertionSort(vector <int>&  a) {
a.insert(a.begin(), INT_MIN);
int n = a.size();
int num = a[n - 1];
for(int i = n-2; i >= 0; --i){
if(a[i] > num){
a[i + 1] = a[i];
output(a);
}else{
a[i + 1] = num;
output(a);
break;
}
}

}
int main(void) {
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;
}

```

### Python

```import sys

def insertionSort1(n, arr):
probe = arr[-1]

for ind in range(len(arr)-2, -1, -1):
if arr[ind] > probe:
arr[ind+1] = arr[ind]
print(" ".join(map(str, arr)))
else:
arr[ind+1] = probe
print(" ".join(map(str, arr)))
break
if arr[0] > probe:
arr[0] = probe
print(" ".join(map(str, arr)))

if __name__ == "__main__":
n = int(input().strip())
arr = list(map(int, input().strip().split(' ')))
insertionSort1(n, arr)
```

### Java

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

public class Solution {

public static void insertIntoSorted(int[] ar) {
int sort = ar[ar.length - 1];
int i;
for (i = ar.length - 2; (i >= 0) && (ar[i] > sort); i--) {
ar[i + 1] = ar[i];
printArray(ar);
}
ar[i + 1] = sort;
printArray(ar);

}

/* Tail starts here */
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int s = in.nextInt();
int[] ar = new int[s];
for(int i=0;i<s;i++){
ar[i]=in.nextInt();
}
insertIntoSorted(ar);
}

private static void printArray(int[] ar) {
for(int n: ar){
System.out.print(n+" ");
}
System.out.println("");
}

}
```

Note: This problem (Insertion Sort) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.