**Counting Sort 1**

### Problem Statement :

Comparison Sorting Quicksort usually has a running time of n x 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 x log( n ) (worst-case) running time, since n x 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. 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

### Solution :

` ````
Solution in C :
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;
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
static void printArray(int[] ar) {
for (int n : ar) {
System.out.print(n + " ");
}
System.out.println("");
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] ar = new int[100];
for (int i = 0; i < n; i++) {
ar[in.nextInt()]++;
}
printArray(ar);
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
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;
}
In Python3 :
N = int(input())
arr = list(map(int, input().split()))
freq = [0]*100
for i in arr:
freq[i] = freq[i]+1
print (' '.join(map(str, freq)))
```

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