Bubble Sort Algorithm in DSA Rust - 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 does the number of steps grow when sorting more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn bubble_sort(arr: &mut [i32]) {
    let n = arr.len();
    for i in 0..n {
        for j in 0..n - i - 1 {
            if arr[j] > arr[j + 1] {
                arr.swap(j, j + 1);
            }
        }
    }
}
    

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 times, inner loop runs decreasing times from n-1 down to 1, so roughly n x n times.

How Execution Grows With Input

As the list size grows, the number of comparisons grows much faster because each item is compared multiple times.

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

Pattern observation: The operations grow roughly by the square of the input size.

Final Time Complexity

Time Complexity: O(n²)

This means if you double the number of items, the time taken roughly quadruples.

Common Mistake

[X] Wrong: "Bubble Sort only takes time proportional to the number of items (O(n)) because it just looks at each item once."

[OK] Correct: Bubble Sort compares pairs many times in nested loops, so it does much more work than just one pass.

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 add a flag to stop early when no swaps happen in a 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. Select the smallest element and move it to the beginning

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

C. Divide the list into halves and sort each half separately

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

Solution

1. Step 1: Understand Bubble Sort process

   Bubble Sort compares adjacent elements and swaps them if they are out of order.

2. Step 2: Compare with other sorting methods

   Other options describe different sorting algorithms like merge sort, selection sort, and insertion sort.

3. Final Answer:

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

4. Quick Check:

   Bubble Sort = swapping adjacent pairs [OK]

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

Common Mistakes:

• Confusing Bubble Sort with Selection Sort
• Thinking it divides the list like Merge Sort
• Assuming it inserts elements like Insertion Sort

2. Which of the following is the correct Rust syntax to swap two elements a[i] and a[i+1] in a vector a during Bubble Sort?

easy

A. std::mem::swap(a[i], a[i + 1]);

B. swap(a[i], a[i + 1]);

C. a.swap(i, i + 1);

D. a[i], a[i + 1] = a[i + 1], a[i];

Solution

1. Step 1: Identify Rust vector swap method

   Rust vectors have a built-in method swap(index1, index2) to swap elements safely.

2. Step 2: Check other options for correctness

   std::mem::swap(a[i], a[i + 1]); lacks mutable references and won't compile, swap(a[i], a[i + 1]); is invalid syntax with no such function, a[i], a[i + 1] = a[i + 1], a[i]; is invalid Python-style syntax in Rust.

3. Final Answer:

   a.swap(i, i + 1); -> Option C

4. Quick Check:

   Rust vector swap method = a.swap(i, i + 1) [OK]

Hint: Use vector.swap(i, j) to swap elements in Rust [OK]

Common Mistakes:

• Using Python or C++ style swap syntax
• Trying to swap values directly without references
• Calling std::mem::swap on vector elements incorrectly

3. What will be the output of the following Rust code snippet after applying one full pass of Bubble Sort on the vector?

let mut a = vec![4, 3, 2, 1];
for i in 0..a.len() - 1 {
    if a[i] > a[i + 1] {
        a.swap(i, i + 1);
    }
}
println!("{:?}", a);

medium

A. [3, 2, 1, 4]

B. [1, 2, 3, 4]

C. [4, 3, 2, 1]

D. [2, 3, 1, 4]

Solution

1. Step 1: Trace the single pass swaps

   Start with [4, 3, 2, 1]. Compare pairs: - 4 > 3: swap -> [3, 4, 2, 1] - 4 > 2: swap -> [3, 2, 4, 1] - 4 > 1: swap -> [3, 2, 1, 4]

2. Step 2: Final vector after one pass

   After one pass, largest element 4 bubbles to the end. Vector is [3, 2, 1, 4].

3. Final Answer:

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

4. Quick Check:

   One pass moves largest to end = [3, 2, 1, 4] [OK]

Hint: One pass moves largest element to the end [OK]

Common Mistakes:

• Assuming full sort after one pass
• Not swapping correctly on each comparison
• Confusing indices and swapping wrong elements

4. Identify the bug in this Rust Bubble Sort code snippet and select the correct fix:

let mut a = vec![5, 1, 4, 2, 8];
for i in 0..a.len() {
    for j in 0..a.len() {
        if a[j] > a[j + 1] {
            a.swap(j, j + 1);
        }
    }
}

medium

A. Change inner loop to 0..a.len() - 1 to avoid out-of-bounds

B. Change outer loop to 0..a.len() - 1 to avoid missing last element

C. Replace a.swap(j, j + 1) with std::mem::swap(&mut a[j], &mut a[j + 1])

D. Add a break statement inside inner loop to stop early

Solution

1. Step 1: Identify out-of-bounds error

   Inner loop uses j in 0..a.len(), so j+1 can be equal to a.len(), causing panic.

2. Step 2: Fix inner loop range

   Changing inner loop to 0..a.len() - 1 prevents accessing index out of range.

3. Final Answer:

   Change inner loop to 0..a.len() - 1 to avoid out-of-bounds -> Option A

4. Quick Check:

   Inner loop must stop before last index [OK]

Hint: Inner loop must run till length-1 to avoid index error [OK]

Common Mistakes:

• Not adjusting inner loop range causing panic
• Changing outer loop instead of inner loop
• Using wrong swap method causing compile error

5. Given a vector vec![3, 0, -1, 5, 2], how many total swaps will Bubble Sort perform to fully sort the vector in ascending order?

hard

A. 4

B. 7

C. 6

D. 5

Solution

1. Step 1: Trace swaps pass by pass

   Initial: [3, 0, -1, 5, 2] Pass 1 swaps: - 3 > 0 swap -> [0, 3, -1, 5, 2] - 3 > -1 swap -> [0, -1, 3, 5, 2] - 5 > 2 swap -> [0, -1, 3, 2, 5] Total swaps: 3

2. Step 2: Continue passes and count swaps

   Pass 2: - 0 > -1 no - -1 > 3 no - 3 > 2 swap -> [0, -1, 2, 3, 5] Total swaps: 1 Pass 3: - 0 > -1 no - -1 > 2 no No swaps Pass 4: No swaps Total swaps overall = 3 + 1 + 0 + 0 = 4 (recheck)

3. Step 3: Recount carefully

   Actually, in Pass 1, 3 > 0 swap (1), 3 > -1 swap (2), 5 > 2 swap (3) Pass 2, 0 > -1 no, -1 > 3 no, 3 > 2 swap (4) Pass 3, 0 > -1 no, -1 > 2 no Pass 4 no swaps Total swaps = 4

4. Step 4: Check for missed swaps

   Pass 2 also compares 2 and 3, no swap; Pass 3 no swaps; Pass 4 no swaps. But initial count missed the swap between 3 and 2 in Pass 2, and also the swap between 3 and 2 in Pass 2 is counted once. Total swaps = 4 as per trace.

5. Step 5: Verify by counting inversions

   Count inversions in initial array: (3,0), (3,-1), (3,2), (0,-1), (5,2) Total inversions = 5 Bubble Sort swaps equal to number of inversions, so total swaps = 5

6. Final Answer:

   5 -> Option D

7. Quick Check:

   Number of swaps equals number of inversions = 5 [OK]

Hint: Count swaps each pass until no swaps remain or count inversions [OK]

Common Mistakes:

• Counting comparisons instead of swaps
• Missing swaps in later passes
• Assuming swaps equal to length minus one