Bubble Sort Algorithm in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bubble Sort on a small array

What is the output of the following bubble sort code on the array [4, 2, 5, 1]?

DSA C

void bubbleSort(int arr[], int n) {
    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;
            }
        }
    }
}

int main() {
    int arr[] = {4, 2, 5, 1};
    int n = 4;
    bubbleSort(arr, n);
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    return 0;
}

A 5 4 2 1

B 1 2 4 5

C 2 1 4 5

D 4 2 1 5

💡 Hint

Bubble sort swaps adjacent elements if they are in wrong order, pushing largest to the end each pass.

The bubble sort compares adjacent elements and swaps them if the left is greater than the right. After all passes, the array is sorted ascending: 1 2 4 5.

❓ Predict Output

intermediate

Number of swaps in bubble sort

How many swaps does bubble sort perform on the array [3, 1, 4, 2]?

DSA C

int bubbleSortCountSwaps(int arr[], int n) {
    int swapCount = 0;
    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;
                swapCount++;
            }
        }
    }
    return swapCount;
}

int main() {
    int arr[] = {3, 1, 4, 2};
    int n = 4;
    int swaps = bubbleSortCountSwaps(arr, n);
    printf("%d", swaps);
    return 0;
}

A 3

B 5

C 4

D 6

💡 Hint

Count each time two adjacent elements are swapped during sorting.

In the first pass: swaps at positions j=0 (3>1) and j=2 (4>2). In the second pass: swap at j=1 (3>2). Total 3 swaps.

🔧 Debug

advanced

Identify the error in this bubble sort code

What error will this bubble sort code cause when run on array [2, 3, 1]?

DSA C

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

A Infinite loop

B No error, sorts correctly

C Array index out of bounds error

D Compilation error due to missing semicolon

💡 Hint

Check the inner loop boundary and the access to arr[j+1].

The inner loop runs j from 0 to n-i, so arr[j+1] accesses arr[n], which is out of bounds causing runtime error.

🧠 Conceptual

advanced

Best case time complexity of bubble sort

What is the best case time complexity of bubble sort when the array is already sorted?

A O(n^2)

B O(log n)

C O(n log n)

D O(n)

💡 Hint

Consider if bubble sort can detect the array is sorted early.

With an optimized bubble sort that stops if no swaps occur, best case is O(n) when array is already sorted.

🚀 Application

expert

Output after partial bubble sort passes

Given array [5, 3, 8, 4, 2], what is the array state after exactly two full passes of bubble sort?

DSA C

void bubbleSortPartial(int arr[], int n, int passes) {
    for (int i = 0; i < passes; 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;
            }
        }
    }
}

int main() {
    int arr[] = {5, 3, 8, 4, 2};
    int n = 5;
    bubbleSortPartial(arr, n, 2);
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    return 0;
}

A 3 4 2 5 8

B 3 5 4 2 8

C 3 4 5 2 8

D 2 3 4 5 8

💡 Hint

Trace each pass carefully, swapping adjacent elements if needed.

After first pass: [3, 5, 4, 2, 8], after second pass: [3, 4, 2, 5, 8].

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