Heap Sort Algorithm in DSA Rust - Time & Space Complexity

Time Complexity: Heap Sort Algorithm

O(n log n)

Understanding Time Complexity

We want to understand how the time taken by heap sort changes as the list size grows.

How does the number of steps increase when sorting bigger lists?

Scenario Under Consideration

Analyze the time complexity of the following heap sort code.

fn heap_sort(arr: &mut [i32]) {
    let n = arr.len();
    for i in (0..n / 2).rev() {
        heapify(arr, n, i);
    }
    for i in (1..n).rev() {
        arr.swap(0, i);
        heapify(arr, i, 0);
    }
}

fn heapify(arr: &mut [i32], n: usize, i: usize) {
    let mut largest = i;
    let left = 2 * i + 1;
    let right = 2 * i + 2;

    if left < n && arr[left] > arr[largest] {
        largest = left;
    }
    if right < n && arr[right] > arr[largest] {
        largest = right;
    }
    if largest != i {
        arr.swap(i, largest);
        heapify(arr, n, largest);
    }
}
    

This code builds a max heap from the array, then repeatedly swaps the largest element to the end and fixes the heap.

Identify Repeating Operations

Look at the loops and recursive calls that repeat work.

• Primary operation: The heapify function called many times to maintain the heap property.
• How many times: heapify runs once for each node when building the heap, then once for each element during sorting, each time possibly recursing down the tree.

How Execution Grows With Input

As the list size grows, the number of operations grows faster than just the list size.

Input Size (n)	Approx. Operations
10	About 30 to 40 heapify steps
100	About 700 to 800 heapify steps
1000	About 10,000 to 12,000 heapify steps

Pattern observation: The operations grow roughly proportional to n times log n, meaning it grows faster than the list size but slower than n squared.

Final Time Complexity

Time Complexity: O(n log n)

This means if you double the list size, the work grows a bit more than double, but not as much as square.

Common Mistake

[X] Wrong: "Heap sort is as slow as bubble sort because both swap elements repeatedly."

[OK] Correct: Heap sort smartly organizes data in a tree structure, reducing the number of swaps and comparisons compared to bubble sort, making it much faster for large lists.

Interview Connect

Understanding heap sort's time complexity shows you can analyze efficient sorting methods, a key skill in many coding challenges and real projects.

Self-Check

"What if we changed the heap from a max heap to a min heap? How would the time complexity change?"

Practice

1. What is the main purpose of the heap data structure in the Heap Sort algorithm?

easy

A. To count the frequency of elements

B. To store elements in sorted order permanently

C. To efficiently find and remove the largest (or smallest) element

D. To reverse the order of elements

Solution

1. Step 1: Understand heap property

   A heap is a special tree where the parent node is always larger (max-heap) or smaller (min-heap) than its children.

2. Step 2: Role in Heap Sort

   Heap Sort uses this property to quickly find the largest element at the root, then remove it and rebuild the heap.

3. Final Answer:

   To efficiently find and remove the largest (or smallest) element -> Option C

4. Quick Check:

   Heap = fast max/min extraction [OK]

Hint: Heap root always holds max/min for quick access [OK]

Common Mistakes:

• Thinking heap stores elements sorted permanently
• Confusing heap with simple array or list
• Assuming heap counts element frequency

2. Which of the following Rust code snippets correctly swaps two elements in a vector during Heap Sort?

easy

A. vec[i] = vec[j];

B. vec.swap(i, j);

C. vec.swap_elements(i, j);

D. vec.exchange(i, j);

Solution

1. Step 1: Identify Rust vector swap method

   Rust's standard Vec<T> has a method called swap that exchanges elements at two indices.

2. Step 2: Check method names

   Only vec.swap(i, j); is valid syntax. Others are incorrect or do not exist.

3. Final Answer:

   vec.swap(i, j); -> Option B

4. Quick Check:

   Rust Vec swap method = vec.swap(i, j) [OK]

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

Common Mistakes:

• Using assignment instead of swap for exchange
• Calling non-existent methods like swap_elements
• Confusing swap with exchange

3. What will be the output vector after applying Heap Sort on vec = vec![4, 10, 3, 5, 1];?

medium

A. [1, 3, 4, 5, 10]

B. [10, 5, 4, 3, 1]

C. [1, 4, 3, 5, 10]

D. [5, 10, 4, 3, 1]

Solution

1. Step 1: Build max heap from vector

   Heap Sort first builds a max heap: [10, 5, 3, 4, 1]

2. Step 2: Extract max and reorder

   Repeatedly swap root with last element and heapify, resulting in sorted ascending vector: [1, 3, 4, 5, 10]

3. Final Answer:

   [1, 3, 4, 5, 10] -> Option A

4. Quick Check:

   Heap Sort output = sorted ascending [OK]

Hint: Heap Sort outputs sorted ascending array [OK]

Common Mistakes:

• Confusing max heap array with sorted output
• Expecting descending order output
• Not heapifying after swaps

4. Identify the bug in this Rust Heap Sort snippet:

fn heapify(vec: &mut Vec<i32>, n: usize, i: usize) {
    let mut largest = i;
    let l = 2 * i + 1;
    let r = 2 * i + 2;
    if l < n && vec[l] > vec[largest] {
        largest = l;
    }
    if r < n && vec[r] > vec[largest] {
        largest = r;
    }
    if largest != i {
        vec.swap(i, largest);
        heapify(vec, n, largest);
    }
}

fn heap_sort(vec: &mut Vec<i32>) {
    let n = vec.len();
    for i in (0..n/2).rev() {
        heapify(vec, n, i);
    }
    for i in (1..n-1).rev() {
        vec.swap(0, i);
        heapify(vec, i, 0);
    }
}

medium

A. The second for loop should start from n-1, not 1

B. The first for loop should be (0..n/2).rev(), which is correct here

C. The heapify call inside second loop uses wrong parameters

D. The swap method is used incorrectly

Solution

1. Step 1: Check second for loop range

   The second loop should run from n-1 down to 1 inclusive to swap root with last element correctly.

2. Step 2: Identify loop range error

   Current loop is (1..n-1).rev(), which iterates from n-2 down to 1, but Rust ranges exclude upper bound, so it misses n-1.

3. Final Answer:

   The second for loop should start from n-1, not 1 -> Option A

4. Quick Check:

   Loop must cover all indices from n-1 down to 1 [OK]

Hint: Second loop must start at n-1 to cover all elements [OK]

Common Mistakes:

• Using wrong loop range excluding last element
• Misunderstanding Rust range bounds
• Incorrect recursive call parameters

5. Given a vector vec = vec![3, 1, 4, 1, 5, 9, 2, 6, 5], how many times will the heapify function be called during the entire Heap Sort process?

hard

A. Approximately 2 * n calls

B. Exactly n calls

C. Exactly log n calls

D. Approximately n * log n calls

Solution

1. Step 1: Understand heapify calls in Heap Sort

   Heapify is called during building the heap (about n/2 calls) and during sorting (about n * log n calls).

2. Step 2: Total calls estimate

   Overall, heapify is called approximately n * log n times because each extraction requires heapify down the tree height (log n).

3. Final Answer:

   Approximately n * log n calls -> Option D

4. Quick Check:

   Heapify calls scale with n log n [OK]

Hint: Heapify calls grow roughly n log n times [OK]

Common Mistakes:

• Assuming heapify calls equal n only
• Thinking heapify calls are constant log n
• Confusing build heap and sort phases