[HackerRank] Counting Sort 1

Interview Preparation Kits > 1 Week Preparation Kit > Day 2 > Counting Sort 1

 

 

Comparison Sorting

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

 

Example

arr  = [1, 1, 3, 2, 1]

All of the values are in the range [0 ... 3], so create an array of zeros, result = [0, 0, 0, 0]. 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]

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

 

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

 

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 result[i] represents the number of times i appeared in arr.

---

#!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'countingSort' function below.
#
# The function is expected to return an INTEGER_ARRAY.
# The function accepts INTEGER_ARRAY arr as parameter.
#

def countingSort(arr):
    # Write your code here
    lst = [0 for i in range(100)]
    for i in arr:
        lst[i] += 1
    print(lst)
    return lst

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input().strip())

    arr = list(map(int, input().rstrip().split()))

    result = countingSort(arr)

    fptr.write(' '.join(map(str, result)))
    fptr.write('\n')

    fptr.close()

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