HackerRank Counting Sort 2 problem solution

In this HackerRank Counting Sort 2 problem you have Given an unsorted list of integers, use the counting sort method to sort the list, and then print the sorted list.

Problem solution in Python programming.

n = int( input() )
ar = [ int(v) for v in input().split() ]

count = [0]*100
for item in ar:
    count[item] += 1

for i in range(100):
    if count[i] > 0:
        print( " ".join( [str(i)]*count[i] ), end=" " )

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static int[] count(int[] ar, int max){
        int[] counts = new int[max+1];
        for(Integer n : ar){
            counts[n] += 1;
        }
        return counts;
    }

    public static void printCount(int[] counts){
        for(int i = 0; i < counts.length; ++i){
            for(int j = 0; j < counts[i]; ++j){
                System.out.print(i + " ");
            }
        }
        System.out.println();
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] ar = new int[n];
        for(int i=0;i<n;i++){
            ar[i]=in.nextInt();
        }
        int[] counts = count(ar, 99);
       // printArray(counts);
        printCount(counts);

    }

}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int n;
    cin>>n;
    const int LEN = 100;
    int dic[LEN] = {0};
    for(int i=0; i<n; i++){
        int tmp;
        cin >> tmp;
        dic[tmp]++;
    }
    for(int i=0; i<LEN; i++){
        for(int j=0; j<dic[i]; j++){
            cout<< i << " ";
        }
    }
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int n;
    scanf("%d", &n);
    int ar1 [n];
    for (int i = 0; i < n; i++) {
        scanf("%d", ar1 + i);
    }
    int ar2 [100];
    for (int j = 0; j < 100; j++) {
        ar2[j] = 0;
    }
    count(ar1, ar2, n);
    for(int k = 0; k < 100; k++) {
        int c = ar2[k];
        while(c > 0) {
            printf("%d ", k);
            c--;
        }
    }
    return 0;
}

void count (int* ar1, int *ar2, int size) {
    for(int i = 0; i < size; i++) {
        ar2[(ar1[i])]++;
    }
}

Problem solution in JavaScript programming.

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input = "";

process.stdin.on('data', function (data) {
    input += data;
});

function sortNumber(a,b) {
    return a - b;
}


function write(text){
    process.stdout.write(""+text+"n");
}
process.stdin.on('end', function () {
    input = input.split("n");
    
    var array = input[1].split(" ");

    for(var a = 0; a < array.length; a++){
        array[a] = parseInt(array[a]);
    }
    
    var ret = array.sort(sortNumber).join(" ");
    write(ret);
});