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Hackerrank Insertion Sort Part 2

Submitted by: @import:stackexchange-codereview··
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Problem

I have started learning Java recently and was looking into the challenges on sorting on Hackerrank.I solved the following problemHackerrank on Insertion Sort.

The challenge was :

In Insertion Sort Part 1, you sorted one element into an array. Using the same approach repeatedly, can you sort an entire unsorted 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 element with just the first element - that is already "sorted" since there's nothing to its left that is smaller.

In this challenge, don't print every time you move an element. Instead, print the array after each iteration of the insertion-sort, i.e., whenever the next element is placed at its correct position.

Since the array composed of just the first element is already "sorted", begin printing from the second element and on.

Input Format
There will be two lines of input:

  • the size of the array



  • a list of numbers that makes up the array



Output Format :
On each line, output the entire array at every iteration.

Sample Input:

6
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


I am trying to get better at writing good code for my solutions..Please give me your suggestions .

import java.util.Scanner;

public class InsertionSortPart2 {

public static void main(String[] args) {

    Scanner keyboard = new Scanner(System.in);
    int size = keyboard.nextInt();

    int[] array = new int[size];

    for (int i = 0; i = 0 && a[j] > key) {
            a[j + 1] = a[j];
            j--;
        }

        a[j + 1] = key;
        for (int f = 0; f < a.length; f++) {
            System.out.print(a[f] + " ");
        }
        System.out.println();
    }

    return a;
  }
}

Solution

Your implementation is quite fine. Some minor improvements are possible.

The insertionsort method doesn't need to return the array. It's not required by the problem, and it's pointless anyway, as the method modifies the input array.

In the main method you called the array array. That's not a very useful name, too generic. Inside insertionsort it's even worse just the letter a. A better name in both places would be for example nums.

The inner while loop would be better as a for loop. The reason is, with a for loop it's easier to not forget to decrement the loop variable:

for (; j >= 0 && nums[j] > key; --j) {


Also, you set j = i - 1, and then you have to remember to use a[j + 1] in two places. It would be slightly simpler to use j = i and rearrange the rest accordingly.

Lastly, the printing of the array really belongs to a separate helper method.

Code Snippets

for (; j >= 0 && nums[j] > key; --j) {

Context

StackExchange Code Review Q#123753, answer score: 2

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