Counting Sort Algorithm in DSA C#

Mental Model

Counting sort counts how many times each number appears, then uses these counts to place numbers in order without comparing them.

Analogy: Imagine sorting colored balls by putting them into labeled boxes for each color, then taking them out in order to get a sorted line.

Unsorted array:
[4] -> [2] -> [2] -> [8] -> [3] -> null

Count array (index = number):
Index: 0 1 2 3 4 5 6 7 8
Count: 0 0 0 0 0 0 0 0 0

Dry Run Walkthrough

Input: array: [4, 2, 2, 8, 3]

Goal: Sort the array in ascending order using counting sort

Step 1: Count how many times each number appears

Count array:
Index: 0 1 2 3 4 5 6 7 8
Count: 0 0 2 1 1 0 0 0 1

Why: We need to know the frequency of each number to place them correctly

Step 2: Modify count array to hold cumulative counts

Count array:
Index: 0 1 2 3 4 5 6 7 8
Count: 0 0 2 3 4 4 4 4 5

Why: Cumulative counts tell us the position of each number in the sorted array

Step 3: Place each element from original array into output array using count positions

Output array after placing elements:
[2] -> [2] -> [3] -> [4] -> [8] -> null

Why: Using counts, we put each number in its correct sorted position

Result:

Sorted array:
2 -> 2 -> 3 -> 4 -> 8 -> null

Annotated Code

DSA C#

using System;

class CountingSort {
    public static void Sort(int[] arr) {
        if (arr.Length == 0) return;

        int max = arr[0];
        for (int i = 1; i < arr.Length; i++) {
            if (arr[i] > max) max = arr[i];
        }

        int[] count = new int[max + 1];

        // Count frequency of each number
        for (int i = 0; i < arr.Length; i++) {
            count[arr[i]]++;
        }

        // Change count[i] so it contains actual position of this number in output
        for (int i = 1; i < count.Length; i++) {
            count[i] += count[i - 1];
        }

        int[] output = new int[arr.Length];

        // Build the output array
        for (int i = arr.Length - 1; i >= 0; i--) {
            output[count[arr[i]] - 1] = arr[i];
            count[arr[i]]--;
        }

        // Copy sorted elements back to original array
        for (int i = 0; i < arr.Length; i++) {
            arr[i] = output[i];
        }
    }

    static void Main() {
        int[] arr = {4, 2, 2, 8, 3};
        Sort(arr);
        for (int i = 0; i < arr.Length; i++) {
            Console.Write(arr[i]);
            if (i != arr.Length - 1) Console.Write(" -> ");
        }
        Console.WriteLine(" -> null");
    }
}

for (int i = 1; i < arr.Length; i++) { if (arr[i] > max) max = arr[i]; }

Find maximum value to size count array

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

Count frequency of each number

for (int i = 1; i < count.Length; i++) { count[i] += count[i - 1]; }

Convert counts to cumulative counts for positions

for (int i = arr.Length - 1; i >= 0; i--) { output[count[arr[i]] - 1] = arr[i]; count[arr[i]]--; }

Place elements in output array from right to left to maintain stability

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

Copy sorted output back to original array

Output

2 -> 2 -> 3 -> 4 -> 8 -> null

Complexity Analysis

Time: O(n + k) because we count frequencies (n) and then process counts up to max value k

Space: O(n + k) because we use output array of size n and count array of size k

vs Alternative: Compared to comparison sorts (O(n log n)), counting sort is faster for small range k but uses extra space

Edge Cases

empty array

function returns immediately without error

DSA C#

if (arr.Length == 0) return;

array with one element

array remains unchanged as it is already sorted

DSA C#

function handles normally without special case

array with all elements the same

count array counts frequency correctly, output is same as input

DSA C#

count[arr[i]]++ line counts frequencies correctly

When to Use This Pattern

When you need to sort integers with a known small range quickly and without comparisons, reach for counting sort because it uses counting and positions to sort in linear time.

Common Mistakes

Mistake: Not using cumulative counts to find correct positions
Fix: Add counts cumulatively so each count[i] holds the position for number i

Mistake: Placing elements from left to right causing unstable sort
Fix: Place elements from right to left to keep original order of equal elements

Mistake: Not copying sorted output back to original array
Fix: Copy output array back to input array after sorting

Summary

Counting sort sorts numbers by counting their occurrences and placing them using cumulative counts.

Use it when sorting integers with a small range efficiently and stably.

The key insight is to use counts to find exact positions without comparing elements.

Practice

1. What is the main idea behind the Counting Sort Algorithm?

easy

A. Count the occurrences of each element and use counts to place elements in order

B. Compare each element with every other element to sort

C. Divide the array into halves and sort recursively

D. Swap adjacent elements if they are in wrong order repeatedly

Solution

1. Step 1: Understand Counting Sort principle

   Counting sort counts how many times each value appears in the input.

2. Step 2: Use counts to place elements

   These counts help to directly place elements in the sorted output without comparisons.

3. Final Answer:

   Count the occurrences of each element and use counts to place elements in order -> Option A

4. Quick Check:

   Counting occurrences = A [OK]

Hint: Counting sort uses frequency counts, not comparisons [OK]

Common Mistakes:

• Thinking counting sort compares elements
• Confusing counting sort with merge sort
• Assuming counting sort works on all data types

2. Which of the following is the correct way to declare the count array for counting sort in C# if the input range is from 0 to 9?

easy

A. int[] count = new int[0];

B. int[] count = new int[10];

C. int[] count = new int[11];

D. int count = new int[9];

Solution

1. Step 1: Determine size of count array

   For range 0 to 9, we need 10 slots to count each number.

2. Step 2: Correct C# syntax for array declaration

   Use int[] count = new int[10]; to create an array of size 10.

3. Final Answer:

   int[] count = new int[10]; -> Option B

4. Quick Check:

   Array size = max value + 1 = 10 [OK]

Hint: Count array size = max element value + 1 [OK]

Common Mistakes:

• Using size less than max value + 1
• Declaring count as single int instead of array
• Using incorrect array size causing index errors

3. What will be the output of the following C# code snippet using counting sort on array int[] arr = {4, 2, 2, 8, 3, 3, 1}?

int max = 8;
int[] count = new int[max + 1];
int[] output = new int[arr.Length];

// Count occurrences
foreach (int num in arr) count[num]++;

// Modify count array
for (int i = 1; i <= max; i++) count[i] += count[i - 1];

// Build output
for (int i = arr.Length - 1; i >= 0; i--) {
    output[count[arr[i]] - 1] = arr[i];
    count[arr[i]]--;
}

// Print output
foreach (int num in output) Console.Write(num + " ");

medium

A. 1 2 2 3 3 4 8

B. 8 4 3 3 2 2 1

C. 1 3 2 2 3 4 8

D. 2 2 3 3 4 8 1

Solution

1. Step 1: Count occurrences of each number

   Count array after counting: [0,1,2,2,1,0,0,0,1]

2. Step 2: Modify count array to cumulative counts

   Cumulative count: [0,1,3,5,6,6,6,6,7]

3. Step 3: Place elements in output using counts

   Placing elements from right to left gives sorted array: 1 2 2 3 3 4 8

4. Final Answer:

   1 2 2 3 3 4 8 -> Option A

5. Quick Check:

   Sorted array = 1 2 2 3 3 4 8 [OK]

Hint: Use cumulative counts to place elements from right [OK]

Common Mistakes:

• Not using cumulative counts before placing elements
• Placing elements from left to right causing wrong order
• Off-by-one errors in indexing output

4. Identify the error in this counting sort code snippet and choose the correct fix:

int max = 5;
int[] count = new int[max];
int[] arr = {1, 4, 3, 2, 5};

foreach (int num in arr) {
    count[num]++;
}

medium

A. Initialize count array with zeros explicitly

B. Change foreach loop to for loop

C. Change count array size to max + 1 to avoid index out of range

D. No error, code is correct

Solution

1. Step 1: Analyze count array size

   Count array size is 5, but max value is 5, so index 5 is out of range.

2. Step 2: Fix by increasing count array size

   Count array should be size max + 1 = 6 to include index 5 safely.

3. Final Answer:

   Change count array size to max + 1 to avoid index out of range -> Option C

4. Quick Check:

   Array size = max + 1 to cover all indices [OK]

Hint: Count array size must be max element + 1 [OK]

Common Mistakes:

• Using count array size equal to max value only
• Ignoring index out of range exceptions
• Assuming default array size is enough

5. You have an array int[] arr = {5, 3, 0, 2, 3, 5, 1, 4} and want to sort it using counting sort. However, the array may contain negative numbers in future. How should you modify the counting sort algorithm to handle negative numbers correctly?

hard

A. Sort negative and positive numbers separately and then merge

B. Ignore negative numbers as counting sort only works for positive numbers

C. Use a dictionary instead of an array for counts

D. Shift all numbers by adding the absolute value of the minimum negative number before counting

Solution

1. Step 1: Identify problem with negative numbers

   Counting sort uses array indices as counts, negative indices are invalid.

2. Step 2: Shift numbers to non-negative range

   Add absolute value of minimum negative number to all elements to make them non-negative.

3. Step 3: Apply counting sort on shifted values

   Sort shifted values, then shift back to original values after sorting.

4. Final Answer:

   Shift all numbers by adding the absolute value of the minimum negative number before counting -> Option D

5. Quick Check:

   Shift negatives to zero or above before counting [OK]

Hint: Shift negatives to zero or above before counting [OK]

Common Mistakes:

• Trying to use negative indices in count array
• Ignoring negative values in input
• Using dictionary without sorting logic