Hackerrank - Counting Sort 1 Solution
2 min read

Hackerrank - Counting Sort 1 Solution

Hackerrank - Counting Sort 1 Solution

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

For example, consider an array . All of the values are in the range , so create an array of zeroes, . 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]

Now we can print the list of occurrences,  or determine the sorted array: .

Challenge
Given a list of integers, count and output the number of times each value appears as a list of space-separated integers.

Function Description

Complete the countingSort function in the editor below. It should return an array of integers where each value is the number of occurrences of the element's index value in the original array.

countingSort has the following parameter(s):

  • arr: an array of integers

Input Format

The first line contains an integer , the number of items in .
Each of the next  lines contains an integer  where .

Constraints


Output Format

Output the number of times every number from  through  appears in  as a list of space-separated integers on one line.

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  represents the number of times  appeared in .

Solution in Python

from collections import Counter

def countingSort(arr):
    s = Counter(arr)
    return (s[i] for i in range(100))

n = int(input())
arr = list(map(int,input().split()))
print(*countingSort(arr))

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