HACKER RANK SOLUTION FOR Counting-Sort-2

HACKER RANK SOLUTION FOR COUNTNG SORT-2

Often, when a list is sorted, the elements being sorted are just keys to other values. For example, if you are sorting files by their size, the sizes need to stay connected to their respective files. You cannot just take the size numbers and output them in order, you need to output all the required file information.

However, if you are not concerned about any other information, then you can simply sort those numbers alone. This makes counting sort very simple. If you already counted the values in the list, you don't need to access the original list again. This challenge involves a simple counting sort where the elements being sorted are all that matter.

Challenge

Given an unsorted list of integers, output the integers in order.

Hint: You can use your previous code that counted the items to print out the actual values in-order.

Input Format 

There will be two lines of input:

n - the size of the list

ar - n space separated numbers that belong to the list

Output Format 

Output all the numbers of the list in-order.

Constraints 

1 <= n <= 1000000

0 <= x < 100 , x ∈ ar

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

1 1 3 3 6 8 9 9 10 12 13 16 16 18 20 21 21 22 23 24 25 25 25 27 27 30 30 32 32 32 33 33 33 34 39 39 40 40 41 42 43 44 44 46 46 48 50 53 56 56 57 59 60 61 63 65 67 67 68 69 69 69 70 70 73 73 74 75 75 76 78 78 79 79 80 81 81 82 83 83 84 85 86 86 87 87 89 89 89 90 90 91 92 94 95 96 98 98 99 99

Explanation

In the output you can see the numbers sorted in ascending order. You can also see that numbers appearing multiple times are printed accordingly.

Solution

#include<stdio.h>

main() 
{
    int n,i,j;
    scanf("%d",&n);
    int ar[n];
    int count[100] = {0}; //for 0 to 100
    for(i=0;i<n;i++)
    {
        scanf("%d",&ar[i]);
        count[ar[i]]++;
    }
    for(i=0;i<100;i++)
    {
        if(count[i]>0)
            {
            for(j=0;j<count[i];j++)
            {
            printf("%d ",i);    
            }
        }
    }
    
}

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