Counting Sort Algorithm in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the count array with zeros.

DSA Java

int[] count = new int[max + 1];
for (int i = 0; i < count.length; i++) {
    count[i] = [1];
}

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing count array with 1 instead of 0.

Using loop variable 'i' as initial value.

The count array must start with zeros to correctly count occurrences.

2fill in blank

medium

Complete the code to count the occurrences of each element in the input array.

DSA Java

for (int i = 0; i < arr.length; i++) {
    count[arr[i]] [1];
}

Drag options to blanks, or click blank then click option'

A--

B+=

C++

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using decrement operator -- instead of increment.

Assigning values instead of incrementing.

We increment the count for each element found in the input array.

3fill in blank

hard

Fix the error in updating the count array to hold cumulative counts.

DSA Java

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

Drag options to blanks, or click blank then click option'

A-

B/

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using multiplication or division which is incorrect here.

Cumulative count is found by adding the previous count to the current count.

4fill in blank

hard

Fill both blanks to place elements into the output array in sorted order.

DSA Java

for (int i = arr.length - 1; i >= 0; i--) {
    output[count[arr[i]] [1]] = arr[i];
    count[arr[i]] [2];
}

Drag options to blanks, or click blank then click option'

A- 1

B+ 1

C--

D++

Attempts:

3 left

💡 Hint

Common Mistakes

Using +1 instead of -1 for index.

Incrementing count instead of decrementing.

We use count[arr[i]] - 1 as index and then decrement count[arr[i]] after placing the element.

5fill in blank

hard

Fill the two blanks to copy the sorted elements back to the original array.

DSA Java

for (int i = 0; i < arr.length; i++) {
    arr[i] = output[[1]];
    [2]++;
}

Drag options to blanks, or click blank then click option'

Ai

Bindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using loop variable i instead of index for output access.

Not incrementing the index variable.

We use an index variable to track position in output, increment it after copying each element.

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 elements directly 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 = main idea [OK]

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

Common Mistakes:

• Thinking Counting Sort compares elements
• Confusing Counting Sort with Quick Sort or Merge Sort
• Assuming Counting Sort works well with all data types

2. Which of the following is the correct way to declare the count array for Counting Sort in Java if the input values range from 0 to 9?

easy

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

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

C. int count = new int[10];

D. int[] count = new int();

Solution

1. Step 1: Understand Java array declaration syntax

   In Java, arrays are declared with type[], name = new type[size];

2. Step 2: Check each option for correct syntax

   int[] count = new int[10]; uses correct syntax: int[] count = new int[10]; int count[] = new int(10); uses parentheses instead of brackets, invalid. int count = new int[10]; misses brackets on left side. int[] count = new int(); misses size in brackets.

3. Final Answer:

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

4. Quick Check:

   Correct Java array syntax = int[] count = new int[10]; [OK]

Hint: Use square brackets and specify size for arrays in Java [OK]

Common Mistakes:

• Using parentheses instead of square brackets
• Not specifying array size when creating
• Incorrect placement of brackets in declaration

3. What will be the output of the following Java code snippet using Counting Sort?

int[] arr = {4, 2, 2, 8, 3, 3, 1};
int max = 8;
int[] count = new int[max + 1];
for (int num : arr) {
    count[num]++;
}
int index = 0;
for (int i = 0; i <= max; i++) {
    while (count[i] > 0) {
        arr[index++] = i;
        count[i]--;
    }
}
System.out.println(java.util.Arrays.toString(arr));

medium

A. [4, 2, 2, 8, 3, 3, 1]

B. [8, 4, 3, 3, 2, 2, 1]

C. [1, 2, 3, 4, 8]

D. [1, 2, 2, 3, 3, 4, 8]

Solution

1. Step 1: Count occurrences of each number

   Count array after loop: count[1]=1, count[2]=2, count[3]=2, count[4]=1, count[8]=1, others=0.

2. Step 2: Rebuild sorted array using counts

   Fill arr with 1 once, 2 twice, 3 twice, 4 once, 8 once in order.

3. Final Answer:

   [1, 2, 2, 3, 3, 4, 8] -> Option D

4. Quick Check:

   Sorted array matches [1, 2, 2, 3, 3, 4, 8] [OK]

Hint: Count frequencies, then overwrite array in order [OK]

Common Mistakes:

• Not initializing count array size correctly
• Forgetting to decrement count during placement
• Printing original array without sorting

4. Identify the error in this Counting Sort code snippet and select the correct fix:

int[] arr = {3, 1, 2};
int max = 3;
int[] count = new int[max];
for (int num : arr) {
    count[num]++;
}

medium

A. Change 'count[num]++' to 'count[num - 1]++' to fix indexing

B. Change 'int[] count = new int[max];' to 'int[] count = new int[max + 1];' to include max index

C. Initialize count array with size 'max - 1' instead of 'max' to save space

D. No error, code is correct as is

Solution

1. Step 1: Check count array size for indexing

   Array 'count' size is max=3, so indices 0,1,2 are valid but input has value 3 which is out of bounds.

2. Step 2: Fix count array size to include max index

   Size should be max+1=4 to allow index 3 access safely.

3. Final Answer:

   Change 'int[] count = new int[max];' to 'int[] count = new int[max + 1];' to include max index -> Option B

4. Quick Check:

   Array size must cover max value index [OK]

Hint: Count array size = max value + 1 to avoid index errors [OK]

Common Mistakes:

• Using count array size equal to max value only
• Adjusting indices incorrectly instead of size
• Ignoring ArrayIndexOutOfBoundsException

5. Given an array arr = {5, 3, 0, 2, 3, 0, 1}, how would you modify the Counting Sort algorithm to sort it if the input can contain negative numbers as well? Choose the best approach.

hard

A. Use a hash map instead of an array for counts

B. Ignore negative numbers as Counting Sort cannot handle them

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

D. Sort only the positive numbers and append negatives at the end

Solution

1. Step 1: Understand Counting Sort limitation with negatives

   Counting Sort uses array indices as counts, so negative indices are invalid.

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

   Find minimum value, add its absolute value to all elements to make them non-negative, then apply Counting Sort.

3. Final Answer:

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

4. Quick Check:

   Shift negatives to positive range for Counting Sort [OK]

Hint: Add offset to handle negatives before counting [OK]

Common Mistakes:

• Trying to use negative indices in count array
• Ignoring negative values in input
• Using incorrect data structures without reason