Question

Consider a binary min-heap containing 105 distinct elements. Let \( k \) be the index (in the underlying array) of the maximum element stored in the heap. The number of possible values of \( k \) is:

• A. 53
• B. 52
• C. 27
• D. 1

Hint:

In a binary heap, the maximum element always resides in the last level. Use the total elements and levels to find its possible indices.

Answer 1

The Correct Option is A

Step 1: Min-Heap Property.
In a binary min-heap, the root (index 1) contains the minimum element, and the elements increase in value as we move down the tree. The maximum element will always be in the last level of the heap. Step 2: Calculating the Last Level.
The number of levels in a binary heap with \( n \) elements is \( \lceil \log_2 n \rceil \). For \( n = 105 \): \[ \lceil \log_2 105 \rceil = 7. \] The 7th level is the last level, and it contains the remaining elements. Step 3: Indices of Last Level Elements.
In the array representation of a binary heap, the indices of the elements in the last level start from \( 64 \) and end at \( 105 \). The maximum element will always be one of these leaf nodes. Step 4: Leaf Nodes in the Array.
The leaf nodes in a binary heap correspond to the indices from \( \lceil n/2 \rceil \) to \( n \). For \( n = 105 \): \[ \lceil 105/2 \rceil = 53. \] Thus, the possible indices of the maximum element range from \( 53 \) to \( 105 \), which gives: \[ 105 - 53 + 1 = 53 \, \text{possible values for } k. \] Final Answer: \[ \boxed{53} \]