# Counting Sort 1 – HackerRank Solution

In this post, we will solve Counting Sort 1 HackerRank Solution. This problem (Counting Sort 1) is a part of HackerRank Problem Solving series.

Contents

## Objective

Comparison Sorting
Quicksort usually has a running time ofÂ n x log(n), but is there an algorithm that can sort even faster? In general, this is not possible. Most sorting algorithms areÂ comparison sorts, i.e. they sort a list just by comparing the elements to one another. A comparison sort algorithm cannot beatÂ n x log(n)Â (worst-case) running time, sinceÂ n x log(n)Â represents the minimum number of comparisons needed to know where to place each element. For more details, you can seeÂ these notesÂ (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 rangeÂ [0 . . . 3], so create an array of zeros,Â result = [0, 0, 0, 0]. The results of each iteration follow:

The frequency array isÂ [0, 3, 1, 1]. These values can be used to create the sorted array as well:Â 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.

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]:Â a frequency array

## Input Format

The first line contains an integerÂ n, the number of items inÂ arr.
Each of the nextÂ nÂ lines contains an integerÂ arr[i]Â whereÂ 0 <= i < n.

## Constraints

• 100 <= n <= 106
• 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 valuesÂ result[i]Â represents the number of timesÂ iÂ appeared inÂ arr.

## Solution – Counting Sort 1 – HackerRank Solution

### C++

```#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

#define TOP_OF_RANGE 100

int main() {
int n; std::cin >> n;
std::vector<int> counts(TOP_OF_RANGE,0);

for (int i = 0,tmp; i < n; i++) {
std::cin >> tmp;
counts[tmp]++;
}

for (auto& a : counts)
std::cout << a << " ";

return 0;
}
```

### Python

```n = int(input())
ar = list(map(int, input().split()))

tot = [0]*100

for j in range(0,n):
temp = ar[j]
tot[temp] += 1
print(*tot, sep =' ')
```

### Java

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

public class Solution {

public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int[] frequencies = new int[100];
for(int i = 0; i < n; i++)
{
int num = input.nextInt();
frequencies[num] = frequencies[num] + 1;
}

for(int i = 0; i < frequencies.length; i++)
{
System.out.print(frequencies[i]+" ");
}
}
}
```

Note: This problem (Counting Sort 1) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.