Comparison Based vs Non Comparison Based Sorting in DSA Java - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform a comparison-based sort using Arrays.sort.

DSA Java

int[] arr = {5, 3, 8, 1};
java.util.Arrays.[1](arr);
System.out.println(java.util.Arrays.toString(arr));

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

Asort

BcopyOf

CbinarySearch

Dfill

Attempts:

3 left

💡 Hint

Common Mistakes

Using copyOf instead of sort

Trying to use fill which fills the array with a value

The Arrays.sort method sorts the array using comparison-based sorting.

2fill in blank

medium

Complete the code to count the frequency of digits using a non-comparison based approach.

DSA Java

int[] digits = {1, 3, 2, 1, 3, 1};
int[] freq = new int[10];
for (int d : digits) {
    freq[[1]]++;
}
System.out.println(java.util.Arrays.toString(freq));

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

Afreq

Bdigits

Cd

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'digits' which is the whole array

Using 'i' which is undefined here

We use the digit value d as the index to increment frequency count.

3fill in blank

hard

Fix the error in the counting sort implementation by filling the correct index for frequency increment.

DSA Java

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

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

Amax

Barr

Ccount

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'arr' which is the whole array

Using 'count' which is the frequency array itself

The variable num holds the current element and is used as index to increment count.

4fill in blank

hard

Fill both blanks to complete the radix sort digit extraction and counting step.

DSA Java

int[] arr = {170, 45, 75, 90, 802, 24, 2, 66};
int exp = 1;
int n = arr.length;
int[] output = new int[n];
int[] count = new int[10];
for (int i = 0; i < n; i++) {
    count[(arr[i] / [1]) % [2]]++;
}

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

Aexp

B10

Cn

Darr.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using array length instead of exponent

Using n instead of 10 for modulo

In radix sort, digits are extracted by dividing by exp and taking modulo 10.

5fill in blank

hard

Fill all three blanks to complete the bucket sort insertion step.

DSA Java

int[] arr = {0, 42, 32, 33, 52, 37, 47, 51};
int n = arr.length;
java.util.List<java.util.List<Integer>> buckets = new java.util.ArrayList<>();
for (int i = 0; i < 10; i++) {
    buckets.add(new java.util.ArrayList<>());
}
for (int num : arr) {
    int index = num / [1];
    buckets.get(index).[2](num);
}
for (int i = 0; i < 10; i++) {
    java.util.Collections.[3](buckets.get(i));
}

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

A10

Badd

Csort

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'remove' instead of 'add'

Using 'sort' as method on list instead of Collections.sort

Bucket index is calculated by dividing by 10, numbers are added to buckets, and each bucket is sorted.

Practice

1. Which of the following sorting algorithms is non-comparison based?

easy

A. Quick Sort

B. Merge Sort

C. Counting Sort

D. Heap Sort

Solution

1. Step 1: Understand comparison vs non-comparison sorting

   Comparison based sorting algorithms decide order by comparing elements. Non-comparison sorts use other methods like counting or indexing.

2. Step 2: Identify Counting Sort type

   Counting Sort counts occurrences of elements and places them directly, without comparing elements.

3. Final Answer:

   Counting Sort -> Option C

4. Quick Check:

   Counting Sort is non-comparison based [OK]

Hint: Counting Sort uses counts, no element comparisons [OK]

Common Mistakes:

• Confusing Merge Sort as non-comparison
• Thinking Quick Sort is non-comparison
• Assuming Heap Sort is non-comparison

2. Which of the following is the correct way to declare a Counting Sort method header in Java?

easy

A. void countingSort(int arr[])

B. public void countingSort(int[] arr)

C. public int countingSort(int[] arr)

D. countingSort(int[] arr) void

Solution

1. Step 1: Recall Java method declaration syntax

   Java methods must specify access modifier, return type, method name, and parameters correctly.

2. Step 2: Check each option

   public void countingSort(int[] arr) uses public access, void return type, and correct parameter syntax. void countingSort(int arr[]) misses access modifier. public int countingSort(int[] arr) returns int incorrectly. countingSort(int[] arr) void has invalid syntax.

3. Final Answer:

   public void countingSort(int[] arr) -> Option B

4. Quick Check:

   Java method header syntax = public void countingSort(int[] arr) [OK]

Hint: Java methods need access modifier and return type first [OK]

Common Mistakes:

• Omitting access modifier
• Wrong return type for void methods
• Incorrect parameter syntax

3. What will be the output of the following Java code snippet using Counting Sort on array [4, 2, 2, 8, 3, 3, 1]?

int[] arr = {4, 2, 2, 8, 3, 3, 1};
countingSort(arr);
System.out.println(Arrays.toString(arr));

// Assume countingSort is correctly implemented

medium

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

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

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

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

Solution

1. Step 1: Understand Counting Sort output

   Counting Sort sorts the array in ascending order by counting occurrences.

2. Step 2: Sort the given array

   Sorted array is [1, 2, 2, 3, 3, 4, 8].

3. Final Answer:

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

4. Quick Check:

   Counting Sort output = sorted array [OK]

Hint: Counting Sort outputs sorted array ascending [OK]

Common Mistakes:

• Confusing original and sorted array
• Assuming reverse order output
• Ignoring duplicates in output

4. Identify the error in the following Counting Sort implementation snippet:

public void countingSort(int[] arr) {
  int max = arr[0];
  for (int i = 1; i <= arr.length; i++) {
    if (arr[i] > max) max = arr[i];
  }
  // rest of the code
}

medium

A. The loop should start from i = 0, not 1.

B. Variable max should be initialized to 0, not arr[0].

C. The if condition should use < max instead of >.

D. Loop condition should be i < arr.length instead of <=.

Solution

1. Step 1: Analyze loop boundary

   Array indices go from 0 to arr.length - 1. Using i <= arr.length causes out-of-bounds error.

2. Step 2: Correct loop condition

   Change loop to i < arr.length to avoid accessing invalid index.

3. Final Answer:

   Loop condition should be i < arr.length instead of <= -> Option D

4. Quick Check:

   Array index out-of-bounds fixed by i < arr.length [OK]

Hint: Array loops use i < length, not <= length [OK]

Common Mistakes:

• Using <= in loop causing runtime error
• Initializing max incorrectly
• Wrong comparison operator in if

5. You want to sort a large array of integers where the values range only from 0 to 1000. Which sorting algorithm is the most efficient choice and why?

hard

A. Counting Sort, because it sorts in O(n + k) time where k is the range.

B. Merge Sort, because it is stable and works well on large data.

C. Quick Sort, because it has average O(n log n) time.

D. Heap Sort, because it uses a heap data structure.

Solution

1. Step 1: Analyze data range and size

   Array values range from 0 to 1000, which is a small fixed range (k = 1001).

2. Step 2: Choose sorting algorithm based on range

   Counting Sort runs in O(n + k), efficient when k is small compared to n, better than comparison sorts O(n log n).

3. Final Answer:

   Counting Sort, because it sorts in O(n + k) time where k is the range. -> Option A

4. Quick Check:

   Small range + large array = Counting Sort best [OK]

Hint: Small range arrays use Counting Sort for best speed [OK]

Common Mistakes:

• Choosing Quick Sort ignoring range advantage
• Assuming Merge Sort always best
• Not considering Counting Sort efficiency