In this post, we will solve HackerRank Insertion Sort – Part 2 Problem Solution.
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.
Example
n = 7
arr = [3, 4, 7, 5, 6, 2, 1]
Working from left to right, we get the following output:
3 4 7 5 6 2 1 3 4 7 5 6 2 1 3 4 5 7 6 2 1 3 4 5 6 7 2 1 2 3 4 5 6 7 1 1 2 3 4 5 6 7
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 an integer, n, the size of arr. The next line contains n space-separated integers arr[i].
Output Format
Print the entire array on a new line at every iteration.
Sample Input
STDIN Function
----- --------
6 n = 6
1 4 3 5 6 2 arr = [1, 4, 3, 5, 6, 2]
Sample Output
1 4 3 5 6 2
1 3 4 5 6 2
1 3 4 5 6 2
1 3 4 5 6 2
1 2 3 4 5 6 Explanation
Skip testing 1 against itself at position 0. It is sorted.
Test position 1 against position 0:4 > 1, no more to check, no change.
Print arr
Test position 2 against positions 1 and 0:
- 3 > 1, so insert 3 at position 1 and move others to the right.
- 3 < 4, new position may be 1. Keep checking.
Print arr
Test position 3 against positions 2, 1, 0 (as necessary): no change.
Print arr
Test position 4 against positions 3, 2, 1, 0: no change.
Print arr
Test position 5 against positions 4, 3, 2, 1, 0, insert 2 at position 1 and move others to the
right.
Print arr
Insertion Sort – Part 2 C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <assert.h>
/* Head ends here */
void insertionSort(int ar_size, int * ar) {
for (int i = 1; i < ar_size; ++i) {
int j = i - 1;
int p = ar[i];
while (j >= 0 && p < ar[j]) {
ar[j+1] = ar[j];
j--;
}
ar[j+1] = p;
printf("%d", ar[0]);
for (int k = 1; k < ar_size; ++k) {
printf(" %d", ar[k]);
}
printf("\n");
}
}
/* Tail starts here */
int main() {
int _ar_size;
scanf("%d", &_ar_size);
int _ar[_ar_size], _ar_i;
for(_ar_i = 0; _ar_i < _ar_size; _ar_i++) {
scanf("%d", &_ar[_ar_i]);
}
insertionSort(_ar_size, _ar);
return 0;
}Insertion Sort – Part 2 C++ Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
void insertionSort(vector<int> &ar) {
for(int i = 1; i < ar.size(); ++i) {
int temp = ar[i];
for(int j = i; j >=0; j--) {
if(ar[j-1] > temp) { ar[j] = ar[j-1]; }
else { ar[j] = temp; break; }
}
for(auto &x : ar) cout << x << " "; cout << endl;
}
}
int main() {
int N; cin >> N;
vector<int> A(N); for(int i = 0; i < N; i++) cin >> A[i];
insertionSort(A);
return 0;
}Insertion Sort – Part 2 C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
class Solution {
/* Head ends here */
static void Main(String[] args)
{
int _ar_size;
_ar_size = Convert.ToInt32(Console.ReadLine());
int[] _ar = new int[_ar_size];
String elements = Console.ReadLine();
String[] split_elements = elements.Split(' ');
for (int _ar_i = 0; _ar_i < _ar_size; _ar_i++)
{
_ar[_ar_i] = Convert.ToInt32(split_elements[_ar_i]);
}
insertionSort(_ar);
}
static void insertionSort(int[] ar)
{
int tmp=0,j,i;
int k, counter = 1;
bool specialCase = false;
if(ar.Length == 1) {specialCase = true; goto print;}
for (k = 0; k < ar.Length - 1; k++)
{
if(counter < ar.Length) tmp = ar[counter];
i = k;
while (i >= 0 && tmp < ar[i])
{
j = ar[i];
ar[i] = tmp;
ar[i + 1] = j;
i--;
}
for (int x = 0; x < ar.Length; x++)
Console.Write("{0} ", ar[x]);
Console.WriteLine();
counter++;
}
print: if (specialCase)
{
ar[0] = tmp;
for (int x = 0; x < ar.Length; x++)
Console.Write("{0} ", ar[x]);
Console.WriteLine();
}
}
}Insertion Sort – Part 2 Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'insertionSort2' function below.
*
* The function accepts following parameters:
* 1. INTEGER n
* 2. INTEGER_ARRAY arr
*/
public static void insertionSort2(int n, List<Integer> arr) {
for (int i=1; i<n; i++){
insertionSort1(i, arr);
print(arr);
}
}
public static void print(List<Integer> arr){
System.out.println(arr.stream().map(String::valueOf).collect(Collectors.joining(" ")));
}
public static void insertionSort1(int n, List<Integer> arr) {
// Write your code here
int val = arr.get(n);
int i = n-1;
while (i >= 0){
int comp = arr.get(i);
if (comp > val){
arr.set(i+1, comp);
}
else {
arr.set(i+1, val);
return;
}
i--;
}
arr.set(0, val);
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(bufferedReader.readLine().trim());
List<Integer> arr = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
.map(Integer::parseInt)
.collect(toList());
Result.insertionSort2(n, arr);
bufferedReader.close();
}
}Insertion Sort – Part 2 JavaScript Solution
function processData(input) {
var lines = input.split("\n");
var N = parseInt(lines.shift());
var list = lines.shift().split(" ");
var i, last, before;
for (i=1; i<list.length; i++){
var j = i;
do {
last = list[j] = parseInt(list[j]);
before = list[j-1] = parseInt(list[j-1]);
if (last < before){
swap(list, j, j-1);
}
j--;
} while (j>0 && last < before)
print(list);
}
}
function print(A){
var str = "";
for (var i=0; i<A.length; i++){
str += A[i] + " ";
}
console.log(str);
}
function swap(A, i1, i2){
var aux = A[i1];
A[i1] = A[i2];
A[i2] = aux;
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});Insertion Sort – Part 2 Python Solution
def insertionSort(ar):
l=len(ar)
val=ar[l-1]
srt=False
N=l-1
while(not srt):
if(N==0 or val>ar[N-1]):
ar[N]=val
srt=True
else:
ar[N]=ar[N-1]
N=N-1
return(ar)
N=int(input())
ar=[int(x,10) for x in input().split()]
for i in range(2,N+1):
ar=insertionSort(ar[:i])+ar[i:]
for x in ar:
print(x,end=' ')
print()Other Solutions
