DATA STRUCTURES/Student Engineering Notebook
Page 07·notes
PRIORITY QUEUES & BINARY HEAPS—~34 MIN READ

Binary Heaps: Invariants, Linear-Time Build-Heap & Heap Sort

“Array indexing, Max-Heapify mechanics, algebraic-geometric proof of O(n) Build-Heap, and priority queue operations.”

1.The Binary Heap as a Compact Array Structure

A Binary Heap is a complete binary tree stored as an array without explicit pointer overhead:

- The tree is completely filled on all levels except possibly the lowest, which is filled from left to right.

- For node at index i (1-indexed):

text{Parent}(i) = lfloor i / 2 rfloor, text{Left}(i) = 2i, text{Right}(i) = 2i + 1.

Heap Property:

- Max-Heap: A[text{Parent}(i)] >= A[i]. The maximum key is always at index 1.

- Min-Heap: A[text{Parent}(i)] <= A[i]. The minimum key is always at index 1.

2.Proof of Linear-Time Build-Heap Complexity

To build a heap from an unordered array of n elements:

Call Max-Heapify bottom-up from index lfloor n/2 rfloor down to 1.

Mathematical Derivation:

- An n-element heap has height lfloor log_2 n rfloor.

- There are at most lceil n / 2^{h+1} rceil nodes at height h.

- Running Max-Heapify on a node of height h takes O(h) time.

Total time:

T(n) <= sum_{h=0}^{lfloor log_2 n rfloor} lceil frac{n}{2^{h+1}} rceil * O(h) <= frac{n}{2} sum_{h=0}^infty frac{h}{2^h}

The infinite arithmetic-geometric series sum_{h=0}^infty frac{h}{2^h} = 2.

Therefore: T(n) <= frac{n}{2} * 2 = O(n).

WORKED EXAM ANALYSIS · STEP-BY-STEP DERIVATIONOriginal Practice Problem · Modeled on GATE Pattern (2 Marks)

3. Worked Problem: Maximum Element Location in Min-Heap

⚠️ Pitfall Warning:

In a Min-Heap, the maximum element is never at an internal node; it is always at a leaf node.

Problem: An array of n = 10 distinct elements forms a Min-Heap. What are the possible indices where the maximum element can reside (using 1-based indexing)?

Step 1: Identify internal nodes and leaves for n = 10:

Last internal node = floor(n / 2) = floor(10 / 2) = 5.

Internal nodes are indices: {1, 2, 3, 4, 5}.

Leaf nodes are indices: {6, 7, 8, 9, 10}.

Step 2: Apply Min-Heap property:

In a Min-Heap, every parent is smaller than both of its children.

If an element is located at any internal node i in {1, 2, 3, 4, 5}, at least one of its children is strictly larger than it.

Therefore, the global maximum CANNOT reside at any internal node.

Step 3: Conclusion:

The maximum element must reside in one of the leaf nodes.

Final Answer: The maximum element can reside at any index in {6, 7, 8, 9, 10}.

SOURCE TRANSPARENCY & CITATIONS

Primary Academic References & Exam Sources

Every concept in Notes From a B.Tech Brain is synthesized from authoritative academic texts, seminal peer-reviewed research, verified exam archives, and renowned university lectures.

Curated University Lecture Series
Introduction to Algorithms (4th Edition) ·
LECTURE SERIES
RECOMMENDED YOUTUBE EDUCATORS & FULL-COURSE PLAYLISTS

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