Heap Extract Min or Max Bubble Down in DSA Rust - Time & Space Complexity

Time Complexity: Heap Extract Min or Max Bubble Down

O(log n)

Understanding Time Complexity

We want to understand how long it takes to remove the smallest or largest item from a heap and fix the heap order.

How does the work grow as the heap gets bigger?

Scenario Under Consideration

Analyze the time complexity of the bubble down step after extracting the root from a heap.

fn bubble_down(heap: &mut Vec<i32>, mut index: usize) {
    let n = heap.len();
    loop {
        let left = 2 * index + 1;
        let right = 2 * index + 2;
        let mut smallest = index;

        if left < n && heap[left] < heap[smallest] {
            smallest = left;
        }
        if right < n && heap[right] < heap[smallest] {
            smallest = right;
        }
        if smallest == index {
            break;
        }
        heap.swap(index, smallest);
        index = smallest;
    }
}

This code moves the root element down the heap to restore the heap property after removal.

Identify Repeating Operations

Look for loops or repeated steps that do the main work.

• Primary operation: Comparing and swapping the current node with its smaller child.
• How many times: At most once per level of the heap, from root down to a leaf.

How Execution Grows With Input

Each step moves down one level in the heap tree. The heap height grows slowly as the heap size grows.

Input Size (n)	Approx. Operations (levels)
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: The number of steps grows slowly, roughly with the height of the heap, which increases as the logarithm of n.

Final Time Complexity

Time Complexity: O(log n)

This means the work to fix the heap grows slowly, only a little more as the heap gets bigger.

Common Mistake

[X] Wrong: "Bubble down takes time proportional to the number of elements in the heap (O(n))."

[OK] Correct: The bubble down only moves down the height of the heap, which is much smaller than the total number of elements.

Interview Connect

Understanding this helps you explain why heaps are efficient for priority queues and sorting, showing you know how data structure operations scale.

Self-Check

"What if the heap was a ternary heap (each node has 3 children) instead of binary? How would the time complexity change?"

Practice

1. What is the main purpose of the bubble down operation in a min-heap after extracting the minimum element?

easy

A. To insert a new element at the bottom of the heap

B. To remove all elements smaller than the root

C. To swap the last element with the root without any further changes

D. To restore the heap property by moving the root element down to its correct position

Solution

1. Step 1: Understand extract-min operation

   Extracting the minimum removes the root, so the last element moves to the root.

2. Step 2: Purpose of bubble down

   Bubble down moves this new root down to restore the min-heap property by swapping with smaller children.

3. Final Answer:

   To restore the heap property by moving the root element down to its correct position -> Option D

4. Quick Check:

   Bubble down fixes heap after extract-min [OK]

Hint: Bubble down fixes heap from root after extract-min [OK]

Common Mistakes:

• Thinking bubble down inserts new elements
• Assuming bubble down removes elements
• Confusing bubble down with bubble up

2. Which of the following Rust function signatures correctly represents a bubble down operation for a min-heap stored in a vector?

easy

A. fn bubble_down(&heap: Vec<i32>, index: usize)

B. fn bubble_down(heap: &mut Vec<i32>, index: usize)

C. fn bubble_down(heap: Vec<i32>, index: usize) -> Vec<i32>

D. fn bubble_down(heap: &Vec<i32>, index: usize)

Solution

1. Step 1: Identify mutability needs

   Bubble down modifies the heap, so it needs a mutable reference to the vector.

2. Step 2: Check parameter types

   fn bubble_down(heap: &mut Vec<i32>, index: usize) uses &mut Vec<i32>, allowing in-place changes without ownership transfer.

3. Final Answer:

   fn bubble_down(heap: &mut Vec<i32>, index: usize) -> Option B

4. Quick Check:

   Mutable reference needed for in-place heap changes [OK]

Hint: Use &mut Vec for in-place heap modification [OK]

Common Mistakes:

• Passing Vec by value causing ownership move
• Using immutable references when mutation is needed
• Incorrect parameter syntax

3. Consider the following Rust code snippet for a min-heap stored in a vector:

let mut heap = vec![1, 3, 6, 5, 9, 8];
heap[0] = heap.pop().unwrap();
bubble_down(&mut heap, 0);
println!("{:?}", heap);

If bubble_down correctly restores the min-heap property, what will be the printed vector?

medium

A. [3, 5, 8, 6, 9]

B. [6, 3, 8, 5, 9]

C. [3, 5, 6, 8, 9]

D. [5, 3, 6, 8, 9]

Solution

1. Step 1: Understand initial heap and pop

   Initial heap: [1,3,6,5,9,8]. After pop(), last element 8 is removed and placed at root: heap[0] = 8, heap = [8,3,6,5,9]

2. Step 2: Bubble down from root

   Compare 8 with children 3 and 6; smallest is 3, swap 8 and 3: [3,8,6,5,9]. Next, bubble down 8 at index 1 with children 5 and 9; smallest is 5, swap: [3,5,6,8,9]. No further swaps needed.

3. Final Answer:

   [3, 5, 6, 8, 9] -> Option C

4. Quick Check:

   Bubble down swaps to restore min-heap [OK]

Hint: Replace root with last, then bubble down to fix heap [OK]

Common Mistakes:

• Not swapping with smallest child
• Stopping bubble down too early
• Incorrect pop and root replacement

4. Given this Rust bubble down function snippet for a max-heap, identify the bug:

fn bubble_down(heap: &mut Vec<i32>, mut index: usize) {
    let len = heap.len();
    loop {
        let left = 2 * index + 1;
        let right = 2 * index + 2;
        let mut largest = index;
        if left < len && heap[left] > heap[largest] {
            largest = left;
        }
        if right < len && heap[right] > heap[largest] {
            largest = right;
        }
        if largest == index {
            break;
        }
        heap.swap(index, largest);
        // Missing update of index here
    }
}

medium

A. The index variable is not updated after swapping, causing an infinite loop

B. The function uses a min-heap condition instead of max-heap

C. The left and right child indices are calculated incorrectly

D. The swap function is used incorrectly

Solution

1. Step 1: Analyze loop behavior

   After swapping, the index must move to the child's position to continue bubbling down.

2. Step 2: Identify missing update

   The code does not update index = largest after swap, so loop repeats with same index causing infinite loop.

3. Final Answer:

   The index variable is not updated after swapping, causing an infinite loop -> Option A

4. Quick Check:

   Update index after swap to continue bubble down [OK]

Hint: Always update index after swap in bubble down loop [OK]

Common Mistakes:

• Forgetting to update index after swap
• Mixing min-heap and max-heap conditions
• Incorrect child index calculations

5. You have a max-heap represented as [50, 30, 40, 10, 20, 35]. After extracting the maximum element (50), you replace the root with the last element and perform bubble down. What is the final heap array after bubble down completes?

hard

A. [40, 30, 35, 10, 20]

B. [40, 30, 10, 20, 35]

C. [35, 30, 40, 10, 20]

D. [35, 30, 20, 10, 40]

Solution

1. Step 1: Replace root with last element

   Remove 50, move last element 35 to root: [35, 30, 40, 10, 20]

2. Step 2: Bubble down from root

   Compare 35 with children 30 and 40; largest is 40, swap: [40, 30, 35, 10, 20]. Next, bubble down 35 at index 2 with children none (leaf), stop.

3. Final Answer:

   [40, 30, 35, 10, 20] -> Option A

4. Quick Check:

   Bubble down swaps root with largest child until heap restored [OK]

Hint: Replace root with last, bubble down swapping with largest child [OK]

Common Mistakes:

• Not swapping with largest child
• Stopping bubble down too early
• Incorrect root replacement