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In this post, you will find the solution for **Counting Sort 1** **in Java-HackerRank Problem**. We are providing the **correct and tested solutions** of coding problems present on **HackerRank**. If you are not able to solve any problem, then you can take help from our Blog/website.

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**Introduction To Algorithm**

The word **Algorithm** means “a process or set of rules to be followed in calculations or other problem-solving operations”. Therefore Algorithm refers to a set of rules/instructions that step-by-step define how a work is to be executed upon in order to get the expected results.

**Advantages of Algorithms:**

- It is easy to understand.
- Algorithm is a step-wise representation of a solution to a given problem.
- In Algorithm the problem is broken down into smaller pieces or steps hence, it is easier for the programmer to convert it into an actual program.

** Link for the Problem** – Counting Sort 1 – Hacker Rank Solution

Counting Sort 1– Hacker Rank Solution

**Problem:**

**Comparison Sorting**

Quicksort usually has a running time of , but is there an algorithm that can sort even faster? In general, this is not possible. Most sorting algorithms are *comparison sorts*, i.e. they sort a list just by comparing the elements to one another. A comparison sort algorithm cannot beat (worst-case) running time, since represents the minimum number of comparisons needed to know where to place each element. For more details, you can see these notes (PDF).

**Alternative Sorting**

Another sorting method, the *counting sort*, does not require comparison. Instead, you create an integer array whose index range covers the entire range of values in your array to sort. Each time a value occurs in the original array, you increment the counter at that index. At the end, run through your counting array, printing the value of each non-zero valued index that number of times.

**Example**

All of the values are in the range , so create an array of zeros, . The results of each iteration follow:

i arr[i] result 0 1 [0, 1, 0, 0] 1 1 [0, 2, 0, 0] 2 3 [0, 2, 0, 1] 3 2 [0, 2, 1, 1] 4 1 [0, 3, 1, 1]

The frequency array is . These values can be used to create the sorted array as well: .

**Note**

For this exercise, always return a frequency array with 100 elements. The example above shows only the first 4 elements, the remainder being zeros.

**Challenge**

Given a list of integers, count and return the number of times each value appears as an array of integers.

**Function Description**

Complete the *countingSort* function in the editor below.

countingSort has the following parameter(s):

*arr[n]:*an array of integers

**Returns**

*int[100]:*a frequency array

**Input Format**

The first line contains an integer , the number of items in .

Each of the next lines contains an integer where .

**Constraints**

**Sample Input**

100 63 25 73 1 98 73 56 84 86 57 16 83 8 25 81 56 9 53 98 67 99 12 83 89 80 91 39 86 76 85 74 39 25 90 59 10 94 32 44 3 89 30 27 79 46 96 27 32 18 21 92 69 81 40 40 34 68 78 24 87 42 69 23 41 78 22 6 90 99 89 50 30 20 1 43 3 70 95 33 46 44 9 69 48 33 60 65 16 82 67 61 32 21 79 75 75 13 87 70 33

**Sample Output**

0 2 0 2 0 0 1 0 1 2 1 0 1 1 0 0 2 0 1 0 1 2 1 1 1 3 0 2 0 0 2 0 3 3 1 0 0 0 0 2 2 1 1 1 2 0 2 0 1 0 1 0 0 1 0 0 2 1 0 1 1 1 0 1 0 1 0 2 1 3 2 0 0 2 1 2 1 0 2 2 1 2 1 2 1 1 2 2 0 3 2 1 1 0 1 1 1 0 2 2

**Explanation**

Each of the resulting values represents the number of times appeared in .

Counting Sort 1 – Hacker Rank Solution

import java.util.Scanner; /** * @author techno-RJ * */ public class CountingSort1 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int a[] = new int[100]; for (int i = 0; i < N; i++) { a[sc.nextInt()]++; } for (int j = 0; j < N; j++) { System.out.print(a[j] + " "); } sc.close(); } }