Counting Sort 1 HackerRank Solution in C, C++, Java, Python

Comparison Sorting

Quicksort usually has a running time of n*log(n), 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 n*log(n) (worst-case) running time, since n*log(n) 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

arr=[1,1,3,2,1]

All of the values are in the range [0…3], so create an array of zeros,result = [0,0,0,0] . 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 [0,3,1,1]. These values can be used to create the sorted array as well: sorted=[1,1,1,2,3].

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 n, the number of items in arr.

Each of the next n lines contains an integer arr[i] where 0<=i<n.

Constraints

100<=n<=10^6

0<=arr[i]<100

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 result[i] represents the number of times i appeared in arr.

 

Counting Sort 1 HackerRank Solution in C

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

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int n,i;
    scanf("%d",&n);
    int b[100],a;
    for(i=0;i<100;i++)
    {   
         b[i]=0;
    }
    for(i=0;i<n;i++)
    {    scanf("%d",&a);
         b[a]++;
    }
    for(i=0;i<100;i++)
    {   
        printf("%d ", b[i]);
    }
   
    return 0;
}

 

Counting Sort 1 HackerRank Solution in C++

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

vector<int>List;
auto main()->int
{
    int size;
    cin >> size;
    //Initialize
    List.resize(100,0);
    for (int i = 0; i != size; i++)
    {
        int n;
        cin >> n;
        List[n] = List[n] + 1;
    }
    for (int r = 0; r != List.size(); r++)
    {
        cout << List[r]<<" ";
    }
    return 0;
}

 

Counting Sort 1 HackerRank Solution in Java

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) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        try{
            BufferedReader buf=new BufferedReader(new InputStreamReader(System.in));
            int n=Integer.parseInt(buf.readLine());
            int[] count=new int[100];
            String[] str=buf.readLine().split(" ");
            for(int i=0;i<n;i++){
                count[Integer.parseInt(str[i])]++;
            }
            for(int i=0;i<99;i++){
                System.out.print(count[i]+" ");
            }
            System.out.println(count[99]);
        }catch(Exception e){}
    }
}

 

Counting Sort 1 HackerRank Solution in Python

def main():
    t = input()
    w = raw_input()
    w = w.split()
    for i in range(100):
        print w.count((str)(i)),
    
if __name__ == '__main__':
    main()

 

Counting Sort 1 HackerRank Solution in C#

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace counting
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.ReadLine();
            int[] res = new int[100];

            foreach (int i in Console.ReadLine().Trim().Split(' ').Select(int.Parse))
            {
                res[i]++;
            }
            foreach (int r in res)
            {
                Console.Write("{0} ",r  );
            }
        }
    }
}

 

Attempt Counting Sort 1 HackerRank Challenge

Link – https://www.hackerrank.com/challenges/countingsort1/

Next HackerRank Challenge Solution 

Link – https://exploringbits.com/counting-sort-2-hackerrank-solution/

 

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