DevTut

# Counting Sort

# Counting Sort Basic Information

Counting sort (opens new window) is an integer sorting algorithm for a collection of objects that sorts according to the keys of the objects.

Steps

  1. Construct a working array C that has size equal to the range of the input array A.
  2. Iterate through A, assigning C[x] based on the number of times x appeared in A.
  3. Transform C into an array where C[x] refers to the number of values ≤ x by iterating through the array, assigning to each C[x] the sum of its prior value and all values in C that come before it.
  4. Iterate backwards through A, placing each value in to a new sorted array B at the index recorded in C. This is done for a given A[x] by assigning B[C[A[x]]] to A[x], and decrementing C[A[x]] in case there were duplicate values in the original unsorted array.

Example of Counting Sort

Counting Sort (opens new window)

Auxiliary Space: O(n+k)
Time Complexity: Worst-case: O(n+k), Best-case: O(n), Average-case O(n+k)

# Psuedocode Implementation

Constraints:

  1. Input (an array to be sorted)
  2. Number of element in input (n)
  3. Keys in the range of 0..k-1 (k)
  4. Count (an array of number)

Pseudocode:

for x in input:
    count[key(x)] += 1
total = 0
for i in range(k):
    oldCount = count[i]
    count[i] = total
    total += oldCount
for x in input:
    output[count[key(x)]] = x
    count[key(x)] += 1
return output

# C# Implementation

public class CountingSort
{
    public static void SortCounting(int[] input, int min, int max)
    {
        var count = new int[max - min + 1];
        var z = 0;

        for (var i = 0; i < count.Length; i++)
            count[i] = 0;

        foreach (int i in input)
            count[i - min]++;

        for (var i = min; i <= max; i++)
        {
            while (count[i - min]-- > 0)
            {
                input[z] = i;
                ++z;
            }
        }
    }

    public static int[] Main(int[] input)
    {
        SortCounting(input, input.Min(), input.Max());
        return input;
    }
}

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