Bubble Sort Algorithm in DSA C - Time & Space Complexity

Time Complexity: Bubble Sort Algorithm

O(n²)

Understanding Time Complexity

We want to understand how the time taken by Bubble Sort changes as the list size grows.

How many steps does it take to sort a list of numbers?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
            }
        }
    }
}
    

This code sorts an array by repeatedly swapping adjacent elements if they are in the wrong order.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing and possibly swapping adjacent elements inside nested loops.
• How many times: Outer loop runs (n-1) times, inner loop runs decreasing times from (n-1) down to 1.

How Execution Grows With Input

As the list size grows, the number of comparisons grows roughly with the square of the size.

Input Size (n)	Approx. Operations
10	About 45 comparisons
100	About 4,950 comparisons
1000	About 499,500 comparisons

Pattern observation: Doubling the input size roughly quadruples the work done.

Final Time Complexity

Time Complexity: O(n²)

This means the time to sort grows quickly as the list gets bigger, roughly proportional to the square of the list size.

Common Mistake

[X] Wrong: "Bubble Sort only takes time proportional to the list size (O(n)) because it just compares neighbors."

[OK] Correct: Because it uses nested loops, it compares many pairs multiple times, making the total steps grow much faster than the list size.

Interview Connect

Understanding Bubble Sort's time complexity helps you appreciate why more efficient sorting methods are used in real projects.

Self-Check

"What if we stop the algorithm early if no swaps happen in a full pass? How would the time complexity change in the best case?"

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