Bubble Sort Algorithm in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the Bubble Sort algorithm?

Bubble Sort repeatedly compares adjacent elements and swaps them if they are in the wrong order, "bubbling" the largest unsorted element to the end in each pass.

beginner

How many passes does Bubble Sort make in the worst case for an array of size n?

In the worst case, Bubble Sort makes n-1 passes over the array to fully sort it.

intermediate

What is the time complexity of Bubble Sort in the average and worst cases?

The time complexity is O(n²) in both average and worst cases because it compares pairs repeatedly for each element.

intermediate

How can Bubble Sort be optimized to stop early if the array is already sorted?

By adding a flag to check if any swaps happened in a pass. If no swaps occur, the array is sorted and the algorithm can stop early.

beginner

Show the state of the array [5, 3, 8, 4, 2] after the first pass of Bubble Sort.

After first pass: [3, 5, 4, 2, 8] because 5 and 3 swapped, 8 stayed, 4 and 2 swapped, and 8 bubbled to the end.

What does Bubble Sort compare during its sorting process?

A Adjacent elements

B First and last elements

C Random elements

D Only the smallest elements

Bubble Sort compares adjacent elements to decide if they need to be swapped.

What is the best case time complexity of Bubble Sort when the array is already sorted?

A O(n)

B O(n log n)

C O(n²)

D O(log n)

With an optimization to stop early, Bubble Sort runs in O(n) time if the array is already sorted.

Which of these is a key disadvantage of Bubble Sort?

A It uses extra memory

B It cannot sort negative numbers

C It only works on sorted arrays

D It is slow for large arrays

Bubble Sort is slow (O(n²)) for large arrays compared to more efficient algorithms.

In Bubble Sort, what happens to the largest element in each pass?

A It moves to the beginning

B It bubbles to the end

C It is removed

D It stays in place

The largest unsorted element moves or "bubbles" to the end of the array in each pass.

How can you tell Bubble Sort is finished sorting the array?

A When the first element is the smallest

B When the array length is zero

C When no swaps happen in a pass

D When the last element is the largest

If no swaps occur during a pass, the array is sorted and the algorithm can stop.

Explain how Bubble Sort works step-by-step on a small array.

Think about how bubbles rise to the surface in water.

Describe one way to optimize Bubble Sort and why it helps.

Consider what happens if the array is already sorted.

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