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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.

Example

arr = [1,1,1,3,2,1]

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

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

Solution

// Convert sample input string into a usable List<int>
string s = "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";
string[] inputStrings = s.Split(' ');
int[] inputArray = Array.ConvertAll<string, int>(inputStrings, s => int.Parse(s));
List<int> arrList = inputArray.ToList();

foreach (var item in arrList)
{
    Console.Write(item + " ");
    Console.WriteLine();
}

List<int> output = countingSort1(arrList);

foreach (var item in output)
{
    Console.Write(item + " ");
}


static List<int> countingSort1(List<int> arr)
{
    int[] result = new int[100];

    // take the input List<int> and break it into chunks of 100
    for (int i = 0; i < arr.Count; i += 100)
    {
        for (int j = i; j < i + 100; j++)
        {
            // the int value is the result array index
            // increment the count at that index
            result[arr[j]]++;
        }
    }

    return result.ToList();
}
```
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  • \$\begingroup\$ @Heslacher thank you for that... Also, just a noob question, in English, what's the meaning of: result[arr[i]]++; on your solution? I find it hard to comprehend. I appreciate it your time. Thank you so much \$\endgroup\$
    – A Y
    Feb 5, 2023 at 5:06

1 Answer 1

3
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Only focusing on countingSort1()

  • using 2 loops doesn't buy you anything, it just blows up your code. But this is dangerous as well if you plan to reuse this method later. Assume you pass in a List whichs Count isn't a multiple of 100, your code will blow up with an ArgumentOutOfRangeException
  • the comment above the first for loop doesn't help anyone because it is stating what is done instead of why something is done.

Taking this into account you could use either

static List<int> countingSort1(List<int> arr)
{
    int[] result = new int[100];

    for(int i = 0; i < arr.Count; i++)
    {
        result[arr[i]]++;
    }

    return result.ToList();
}

or slightly shorter

static List<int> countingSort1(List<int> arr)
{
    int[] result = new int[100];

    arr.ForEach(a => result[a]++);

    return result.ToList();
}
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  • \$\begingroup\$ Thank you for your code review. Your solution is similar to my initial solution for the input shown above. It worked for that input and any input that does not exceed n = 100. However, when I submitted my solution for much larger hacker rank data sets of up to n = 1000, it failed. My solution is correct for the problem statement above including the constraints given. My solution may not be a good generalized counting sort algorithm. A more generalized solution would be better. For now, this solution solves the HackerRank Counting Sort 1 challenge for all HackerRank test cases. \$\endgroup\$ Mar 17, 2022 at 15:41

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