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In this post, you will find the solution for **The Full Counting Sort 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** – The Full Counting Sort– Hacker Rank Solution

The Full Counting Sort– Hacker Rank Solution

**Problem:**

Use the counting sort to order a list of strings associated with integers. If two strings are associated with the same integer, they must be printed in their original order, i.e. your sorting algorithm should be *stable*. There is one other twist: strings in the first half of the array are to be replaced with the character `-`

(dash, ascii 45 decimal).

Insertion Sort and the simple version of Quicksort are stable, but the faster in-place version of Quicksort is not since it scrambles around elements while sorting.

Design your counting sort to be stable.

**Example**

The first two strings are replaced with ‘-‘. Since the maximum associated integer is , set up a helper array with at least two empty arrays as elements. The following shows the insertions into an array of three empty arrays.

i string converted list 0 [[],[],[]] 1 a - [[-],[],[]] 2 b - [[-],[-],[]] 3 c [[-,c],[-],[]] 4 d [[-,c],[-,d],[]]

The result is then printed: .

**Function Description**

Complete the *countSort* function in the editor below. It should construct and print the sorted strings.

countSort has the following parameter(s):

*string arr[n][2]:*each*arr[i]*is comprised of two strings,*x*and*s*

**Returns**

– Print the finished array with each element separated by a single space.

**Note**: The first element of each , , must be cast as an integer to perform the sort.

**Input Format**

The first line contains , the number of integer/string pairs in the array .

Each of the next contains and , the integers (as strings) with their associated strings.

**Constraints**

is even

consists of characters in the range

**Output Format**

Print the strings in their correct order, space-separated on one line.

**Sample Input**

20 0 ab 6 cd 0 ef 6 gh 4 ij 0 ab 6 cd 0 ef 6 gh 0 ij 4 that 3 be 0 to 1 be 5 question 1 or 2 not 4 is 2 to 4 the

**Sample Output**

- - - - - to be or not to be - that is the question - - - -

**Explanation**

The correct order is shown below. In the array at the bottom, strings from the first half of the original array were replaced with dashes.

0 ab 0 ef 0 ab 0 ef 0 ij 0 to 1 be 1 or 2 not 2 to 3 be 4 ij 4 that 4 is 4 the 5 question 6 cd 6 gh 6 cd 6 gh

sorted = [['-', '-', '-', '-', '-', 'to'], ['be', 'or'], ['not', 'to'], ['be'], ['-', 'that', 'is', 'the'], ['question'], ['-', '-', '-', '-'], [], [], [], []]

The Full Counting Sort – Hacker Rank Solution

import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) throws Exception { BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); int n = Integer.parseInt(in.readLine()); StringBuffer[] map = new StringBuffer[100]; for(int i = 0; i < 100; i++) { map[i] = new StringBuffer(); } for(int i = 0; i < n; i++) { StringTokenizer tok = new StringTokenizer(in.readLine()); int v = Integer.parseInt(tok.nextToken()); String s = tok.nextToken(); map[v].append(i < n / 2 ? "-" : s).append(" "); } for(int i = 0; i < 100; i++) { System.out.print(map[i]); } System.out.println(); } }