# 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)))```
```

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

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## Insert a Node at the head of a Linked List

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## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e