Build Heap from Array Heapify in DSA Rust - Time & Space Complexity

Time Complexity: Build Heap from Array Heapify

O(n)

Understanding Time Complexity

We want to understand how long it takes to build a heap from an array using heapify. This helps us see how the work grows as the array gets bigger.

What question are we trying to answer? How does the time needed change when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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);
    }
}

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

This code builds a max heap from an unsorted array by calling heapify from the bottom up.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The heapify function, which may call itself recursively to fix the heap property.
• How many times: The build_heap function calls heapify for each non-leaf node, about n/2 times.

How Execution Grows With Input

Heapify runs faster near the bottom of the tree and slower near the top. So, the total work is less than doing full heapify n/2 times.

Input Size (n)	Approx. Operations
10	About 30 operations
100	About 500 operations
1000	About 8000 operations

Pattern observation: The total operations grow a bit faster than n but much slower than n squared.

Final Time Complexity

Time Complexity: O(n)

This means building a heap from an array takes time proportional to the size of the array, growing linearly as the array grows.

Common Mistake

[X] Wrong: "Heapify runs in O(n log n) because it calls heapify on n/2 nodes and each heapify takes O(log n)."

[OK] Correct: Not all heapify calls take log n time. Many are near the bottom and run faster. The total work adds up to O(n), not O(n log n).

Interview Connect

Knowing that building a heap is O(n) helps you explain why heap sort is efficient. It shows you understand how recursive work adds up in tree structures.

Self-Check

"What if we changed the heapify to a non-recursive version? How would the time complexity change?"

Practice

1. What is the main purpose of the heapify process when building a heap from an array?

easy

A. To find the maximum element in the array

B. To sort the array in ascending order

C. To reverse the array elements

D. To rearrange the array elements to satisfy the heap property

Solution

1. Step 1: Understand heap property

   The heap property requires each parent node to be greater (max-heap) or smaller (min-heap) than its children.

2. Step 2: Role of heapify

   Heapify rearranges elements so this property holds for every subtree, starting from the bottom.

3. Final Answer:

   To rearrange the array elements to satisfy the heap property -> Option D

4. Quick Check:

   Heapify = Rearrange for heap property [OK]

Hint: Heapify fixes subtree order to build a heap fast [OK]

Common Mistakes:

• Thinking heapify sorts the array
• Confusing heapify with finding max/min element
• Assuming heapify reverses the array

2. Which of the following Rust code snippets correctly calls the heapify function on an array slice arr starting at index i with heap size n?

easy

A. heapify(arr, i, n);

B. heapify(&mut arr, i, n);

C. heapify(&arr, i, n);

D. heapify(arr, &i, &n);

Solution

1. Step 1: Understand Rust borrowing rules

   To modify an array in Rust, pass a mutable reference: &mut arr.

2. Step 2: Check parameter types

   Indices i and n are usually copied values, so pass by value.

3. Final Answer:

   heapify(&mut arr, i, n); -> Option B

4. Quick Check:

   Mutable reference for array + values for indices [OK]

Hint: Use &mut for arrays when modifying in Rust [OK]

Common Mistakes:

• Passing array without &mut causes compile error
• Passing immutable reference &arr when mutation needed
• Passing references for integers unnecessarily

3. Given the Rust code below, what will be the state of the array after calling build_heap(&mut arr)? Assume build_heap calls heapify correctly to build a max-heap.

fn main() {
    let mut arr = [3, 9, 2, 1, 4, 5];
    build_heap(&mut arr);
    println!("{:?}", arr);
}

medium

A. [9, 4, 5, 1, 3, 2]

B. [5, 4, 3, 2, 1, 9]

C. [9, 3, 5, 1, 4, 2]

D. [3, 9, 2, 1, 4, 5]

Solution

1. Step 1: Understand build_heap process

   Build_heap calls heapify from last non-leaf node upwards to root, fixing subtree heaps.

2. Step 2: Apply heapify on array [3,9,2,1,4,5]

   Heapify rearranges to max-heap: root is largest, children smaller. Result is [9,4,5,1,3,2].

3. Final Answer:

   [9, 4, 5, 1, 3, 2] -> Option A

4. Quick Check:

   Max-heap root 9, children smaller [OK]

Hint: Build heap from bottom up to fix subtrees [OK]

Common Mistakes:

• Assuming array is sorted after build_heap
• Confusing max-heap with min-heap output
• Not applying heapify from last non-leaf node

4. Consider this Rust function snippet for heapify:

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

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

What is the bug in this code that prevents building a max-heap correctly?

medium

A. The recursive call to heapify uses wrong parameter order

B. The parameters n and i are swapped in the function signature

C. The calculation of left and right child indices is incorrect

D. The function does not check if i is within bounds before accessing arr[i]

Solution

1. Step 1: Examine function signature

   The signature is fn heapify(arr: &mut [i32], i: usize, n: usize).

2. Step 2: Check recursive call

   The recursive call is heapify(arr, n, largest);, which passes n as i and largest as n: wrong parameter order.

3. Final Answer:

   The recursive call to heapify uses wrong parameter order -> Option A

4. Quick Check:

   Recursive call must match signature: arr, i, n [OK]

Hint: Check recursive call matches function parameter order exactly [OK]

Common Mistakes:

• Swapping n and i in recursive call
• Miscalculating child indices
• Ignoring bounds checks for indices

5. You want to build a min-heap from the array [7, 2, 6, 3, 9, 1] using the standard heapify approach. Which modification should you make to the heapify function to correctly build a min-heap instead of a max-heap?

hard

A. Call heapify starting from the root instead of last non-leaf node

B. Swap the left and right child index calculations

C. Change all '>' comparisons to '<' when comparing child and parent values

D. Reverse the array before calling heapify

Solution

1. Step 1: Understand heap property difference

   Max-heap requires parent > children; min-heap requires parent < children.

2. Step 2: Modify comparison operators

   To build min-heap, change comparisons from '>' to '<' so smaller values rise to the top.

3. Final Answer:

   Change all '>' comparisons to '<' when comparing child and parent values -> Option C

4. Quick Check:

   Min-heap uses less-than comparisons [OK]

Hint: Flip comparison signs to switch max-heap to min-heap [OK]

Common Mistakes:

• Swapping child indices does not affect heap type
• Reversing array does not build min-heap
• Starting heapify from root breaks heap building