Bubble Sort Algorithm in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap two integers using a temporary variable.

DSA C

void swap(int *a, int *b) {
    int temp = *a;
    *a = [1];
    *b = temp;
}

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

Aa

B*a

Ctemp

D*b

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning the temporary variable instead of the value pointed by b.

Using the pointer variable instead of dereferenced value.

The value pointed by b is assigned to the location pointed by a to complete the swap.

2fill in blank

medium

Complete the condition to continue the outer loop for bubble sort.

DSA C

for (int i = 0; i [1] n - 1; i++) {
    // inner loop
}

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

A<

B>

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or > which causes the loop to run too many times.

Using <= which runs one extra iteration.

The outer loop runs from 0 up to n-2, so the condition is i < n - 1.

3fill in blank

hard

Fix the error in the inner loop condition to avoid out-of-bound access.

DSA C

for (int j = 0; j [1] n - i - 1; j++) {
    if (arr[j] > arr[j + 1]) {
        swap(&arr[j], &arr[j + 1]);
    }
}

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

A>=

B<

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes the loop to access beyond array size.

Using > or >= skips necessary comparisons.

The inner loop must run while j < n - i - 1 to avoid accessing arr[j + 1] out of bounds.

4fill in blank

hard

Fill both blanks to complete the bubble sort nested loops correctly.

DSA C

for (int i = 0; i [1] n - 1; i++) {
    for (int j = 0; j [2] n - i - 1; j++) {
        if (arr[j] > arr[j + 1]) {
            swap(&arr[j], &arr[j + 1]);
        }
    }
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or <= causes index errors.

Mixing operators between loops causes inconsistent behavior.

Both loops use the less than < operator to iterate correctly within array bounds.

5fill in blank

hard

Fill all three blanks to complete the bubble sort function that sorts an array in ascending order.

DSA C

void bubbleSort(int arr[], int n) {
    for (int i = 0; i [1] n - 1; i++) {
        for (int j = 0; j [2] n - i - 1; j++) {
            if (arr[j] [3] arr[j + 1]) {
                swap(&arr[j], &arr[j + 1]);
            }
        }
    }
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or <= in loops causing index errors.

Using < or <= in comparison causing wrong sorting order.

The outer and inner loops use < to stay within bounds, and the comparison uses > to swap when the left element is greater than the right.

Practice

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

easy

A. Use a pivot to partition the array into smaller parts

B. Divide the array into halves and sort each half separately

C. Repeatedly swap adjacent elements if they are in the wrong order

D. Insert each element into its correct position in a sorted sublist

Solution

1. Step 1: Understand the sorting method

   Bubble sort works by comparing adjacent elements and swapping them if they are out of order.

2. Step 2: Identify the correct description

   Repeatedly swap adjacent elements if they are in the wrong order describes this swapping of neighbors repeatedly until sorted.

3. Final Answer:

   Repeatedly swap adjacent elements if they are in the wrong order -> Option C

4. Quick Check:

   Bubble sort = swapping neighbors [OK]

Hint: Bubble sort swaps neighbors repeatedly until sorted [OK]

Common Mistakes:

• Confusing bubble sort with divide and conquer methods
• Thinking bubble sort uses a pivot element
• Mixing bubble sort with insertion sort logic

2. Which of the following is the correct syntax for the inner loop of bubble sort in C to compare adjacent elements in an array arr of size n?

easy

A. for (int j = 0; j <= n - i; j++)

B. for (int j = 0; j < n - i - 1; j++)

C. for (int j = 1; j < n; j++)

D. for (int j = 0; j < n; j++)

Solution

1. Step 1: Understand the inner loop range

   The inner loop runs from 0 to n - i - 1 because after each pass, the last i elements are sorted.

2. Step 2: Match the correct loop syntax

   for (int j = 0; j < n - i - 1; j++) correctly uses for (int j = 0; j < n - i - 1; j++) to avoid out-of-bound access.

3. Final Answer:

   for (int j = 0; j < n - i - 1; j++) -> Option B

4. Quick Check:

   Inner loop limit = n - i - 1 [OK]

Hint: Inner loop stops before last sorted elements [OK]

Common Mistakes:

• Using j < n causes out-of-bound access
• Starting j from 1 misses first element comparison
• Using <= instead of < causes extra iteration

3. What will be the output of the following C code snippet?

int arr[] = {4, 2, 3};
int n = 3;
for (int i = 0; i < n - 1; i++) {
  for (int j = 0; j < n - i - 1; j++) {
    if (arr[j] > arr[j + 1]) {
      int temp = arr[j];
      arr[j] = arr[j + 1];
      arr[j + 1] = temp;
    }
  }
}
for (int i = 0; i < n; i++) {
  printf("%d ", arr[i]);
}

medium

A. 2 4 3

B. 4 3 2

C. 3 2 4

D. 2 3 4

Solution

1. Step 1: Trace first pass of bubble sort

   Compare 4 and 2, swap -> arr = {2, 4, 3}; then compare 4 and 3, swap -> arr = {2, 3, 4}

2. Step 2: Trace second pass and final array

   Second pass compares 2 and 3 (no swap), 3 and 4 (no swap). Array remains {2, 3, 4}.

3. Final Answer:

   2 3 4 -> Option D

4. Quick Check:

   Sorted array = 2 3 4 [OK]

Hint: Bubble sort sorts array ascending by swapping neighbors [OK]

Common Mistakes:

• Not swapping correctly on first pass
• Printing unsorted array
• Confusing loop limits causing partial sort

4. Identify the error in the following bubble sort code snippet and select the correct fix:

void bubbleSort(int arr[], int n) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n - 1; j++) {
      if (arr[j] > arr[j + 1]) {
        int temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }
}

medium

A. Change outer loop to for (int i = 0; i < n - 1; i++)

B. Change inner loop to for (int j = 1; j < n; j++)

C. Remove the swap inside the if condition

D. Change outer loop to for (int i = 1; i < n; i++)

Solution

1. Step 1: Analyze loop ranges for bubble sort

   The outer loop should be for (int i = 0; i < n - 1; i++) because after n-1 passes, array is sorted, avoiding an unnecessary pass.

2. Step 2: Identify the fix

   Change outer loop to for (int i = 0; i < n - 1; i++) correctly optimizes by limiting to n-1 passes.

3. Final Answer:

   Change outer loop to for (int i = 0; i < n - 1; i++) -> Option A

4. Quick Check:

   Outer loop limit = n - 1 [OK]

Hint: Outer loop runs n-1 times in bubble sort [OK]

Common Mistakes:

• Running outer loop n times wastes time
• Changing inner loop start incorrectly
• Removing swap breaks sorting logic

5. Given an array arr = {5, 1, 4, 2, 8}, how many total swaps will bubble sort perform to sort it in ascending order?

hard

A. 4

B. 5

C. 6

D. 7

Solution

1. Step 1: Trace pass 1 (i=0, j < 4)

   Initial: {5,1,4,2,8}
   j=0: 5>1 swap -> {1,5,4,2,8} (1)
   j=1: 5>4 swap -> {1,4,5,2,8} (2)
   j=2: 5>2 swap -> {1,4,2,5,8} (3)
   j=3: 5<8 no -> 3 swaps, array: {1,4,2,5,8}

2. Step 2: Trace pass 2 (i=1, j < 3)

   j=0: 1<4 no
   j=1: 4>2 swap -> {1,2,4,5,8} (4)
   j=2: 4<5 no -> 1 swap, total 4, array: {1,2,4,5,8}

3. Step 3: Trace pass 3 (i=2, j < 2)

   j=0: 1<2 no
   j=1: 2<4 no -> 0 swaps, total 4

4. Step 4: Trace pass 4 (i=3, j < 1)

   j=0: 1<2 no -> 0 swaps, total 4

5. Step 5: Total swaps counted = 4

   But we missed one swap in pass 2: Actually, pass 2 has 1 swap (4>2), pass 3 has 0 swaps, pass 4 has 0 swaps. Total swaps = 4.

6. Recounting carefully:

   Pass 1: 3 swaps
   Pass 2: 1 swap
   Pass 3: 0 swaps
   Pass 4: 0 swaps
   Total = 4 swaps

7. Final Answer:

   4 -> Option A

8. Quick Check:

   Total swaps counted = 4 [OK]

Hint: Count swaps each pass carefully to avoid overcounting [OK]

Common Mistakes:

• Counting non-swaps as swaps
• Assuming max swaps per pass without checking
• Forgetting that last elements get sorted early