Counting Sort Algorithm in DSA C++ - Build from Scratch

Counting Sort Algorithm

📖 Scenario: You work in a warehouse where packages are labeled with numbers from 0 to 9. You want to sort these package labels quickly to organize them for delivery.

🎯 Goal: Build a program that uses the Counting Sort Algorithm to sort a list of package labels (numbers from 0 to 9) efficiently.

📋 What You'll Learn

Create an array called packages with the exact values: 4, 2, 2, 8, 3, 3, 1

Create a count array called count with size 10 initialized to 0

Implement the counting sort logic to fill the count array with frequencies of each number in packages

Use the count array to reconstruct the sorted packages array

Print the sorted packages array separated by spaces

💡 Why This Matters

🌍 Real World

Counting sort is useful in sorting items with small ranges quickly, like sorting grades, ages, or package labels.

💼 Career

Understanding counting sort helps in optimizing sorting tasks in software development and data processing roles.

1

Create the initial array of package labels

Create an integer array called packages with these exact values: 4, 2, 2, 8, 3, 3, 1

DSA C++

# Your code here

Hint

Use curly braces {} to list the numbers inside the array.

2

Create the count array to store frequencies

Create an integer array called count of size 10 and initialize all elements to 0

DSA C++

int packages[] = {4, 2, 2, 8, 3, 3, 1};
// Your code here

Hint

Use {0} to initialize all elements to zero.

3

Count the frequency of each number in packages

Use a for loop with variable i from 0 to 6 to increment count[packages[i]] by 1

DSA C++

int packages[] = {4, 2, 2, 8, 3, 3, 1};
int count[10] = {0};
// Your code here

Hint

Loop through each element in packages and increase the count for that number.

4

Rebuild the sorted array and print it

Use two for loops: the outer loop with variable num from 0 to 9, and the inner loop with variable j from 0 to count[num] to overwrite packages with sorted numbers. Then print the sorted packages separated by spaces.

DSA C++

int packages[] = {4, 2, 2, 8, 3, 3, 1};
int count[10] = {0};
for (int i = 0; i < 7; i++) {
    count[packages[i]]++;
}
// Your code here

Hint

Use nested loops to fill packages with sorted numbers and then print them with spaces.

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 place each element directly into 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 = Count the occurrences of each element and use counts to place elements in order [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 any data type

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

easy

A. int count[9] = {0};

B. int count[10] = {0};

C. int count[11] = {0};

D. int count[] = {0};

Solution

1. Step 1: Determine size of count array

   Input values range from 0 to 9, so we need 10 slots (0 through 9).

2. Step 2: Declare count array with size 10

   Using int count[10] = {0}; initializes all counts to zero for indices 0 to 9.

3. Final Answer:

   int count[10] = {0}; -> 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 9 instead of 10
• Not initializing count array to zero
• Declaring count array without size

3. What is the output of the following C++ code snippet using counting sort on array arr = {4, 2, 2, 8, 3, 3, 1}?

void countingSort(int arr[], int n) {
    int output[n];
    int count[9] = {0};
    for (int i = 0; i < n; i++)
        count[arr[i]]++;
    for (int i = 1; i < 9; i++)
        count[i] += count[i - 1];
    for (int i = n - 1; i >= 0; i--) {
        output[count[arr[i]] - 1] = arr[i];
        count[arr[i]]--;
    }
    for (int i = 0; i < n; i++)
        arr[i] = output[i];
}

int main() {
    int arr[] = {4, 2, 2, 8, 3, 3, 1};
    int n = sizeof(arr) / sizeof(arr[0]);
    countingSort(arr, n);
    for (int i = 0; i < n; i++)
        std::cout << arr[i] << " ";
    return 0;
}

medium

A. 2 2 3 3 4 8 1

B. 8 4 3 3 2 2 1

C. 1 2 3 4 8 2 3

D. 1 2 2 3 3 4 8

Solution

1. Step 1: Count frequencies and cumulative counts

   Count array after counting: count[1]=1, count[2]=2, count[3]=2, count[4]=1, count[8]=1; cumulative sums give positions.

2. Step 2: Place elements in output using counts

   Elements placed in output array in sorted order: 1 2 2 3 3 4 8.

3. Final Answer:

   1 2 2 3 3 4 8 -> Option D

4. Quick Check:

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

Hint: Counting sort output is sorted input array [OK]

Common Mistakes:

• Not using cumulative counts correctly
• Placing elements in wrong order
• Using wrong count array size

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

void countingSort(int arr[], int n) {
    int output[n];
    int count[10] = {0};
    for (int i = 0; i <= n; i++)
        count[arr[i]]++;
    for (int i = 1; i < 10; i++)
        count[i] += count[i - 1];
    for (int i = n - 1; i >= 0; i--) {
        output[count[arr[i]] - 1] = arr[i];
        count[arr[i]]--;
    }
    for (int i = 0; i < n; i++)
        arr[i] = output[i];
}

medium

A. Initialize output array with zeros

B. Increase count array size to 11

C. Change loop to for (int i = 0; i < n; i++) in first for loop

D. Change second for loop to start from i = 0

Solution

1. Step 1: Check loop bounds in counting frequencies

   The first for loop uses i <= n, which accesses arr[n] out of bounds (indexing error).

2. Step 2: Fix loop condition to i < n

   Changing to i < n prevents out-of-bounds access and correctly counts frequencies.

3. Final Answer:

   Change loop to for (int i = 0; i < n; i++) in first for loop -> Option C

4. Quick Check:

   Loop bounds must be less than n to avoid errors [OK]

Hint: Loop over array indices from 0 to n-1 [OK]

Common Mistakes:

• Using <= n instead of < n in loops
• Ignoring array out-of-bounds errors
• Changing unrelated loops unnecessarily

5. Given an array arr = {5, 3, 0, 2, 3, 5, 1, 4} with values ranging from 0 to 5, how would you modify the counting sort algorithm to sort this array in descending order?

hard

A. After computing cumulative counts, iterate from highest to lowest index to place elements in output

B. Reverse the input array before sorting with normal counting sort

C. Use a max heap instead of counting sort for descending order

D. Sort normally and then reverse the output array after counting sort completes

Solution

1. Step 1: Understand normal counting sort placement

   Normal counting sort places elements from lowest to highest using cumulative counts.

2. Step 2: Modify placement to iterate from highest to lowest

   To sort descending, iterate counts from max to min and place elements accordingly in output.

3. Final Answer:

   After computing cumulative counts, iterate from highest to lowest index to place elements in output -> Option A

4. Quick Check:

   Reverse iteration in placement = descending sort [OK]

Hint: Reverse placement loop for descending order [OK]

Common Mistakes:

• Just reversing input without changing counting logic
• Using other sorting algorithms instead of modifying counting sort
• Reversing output after sorting without adjusting counts