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

In a binary min-heap, every parent node has a value smaller than or equal to the values of its children. Therefore, the maximum element cannot be an internal node (since an internal node must be smaller than its children) and must reside at one of the leaf nodes.

For a complete binary tree/heap with $n = 105$ elements stored in an array using 1-based indexing:

• The internal nodes are located at indices $1, 2, \dots, \lfloor n/2 \rfloor = 1, 2, \dots, \lfloor 105/2 \rfloor = 52$.

• The leaf nodes are located at indices from $\lfloor n/2 \rfloor + 1$ to $n$, which is from index $53$ to index $105$.

The total number of leaf nodes is:

Since the maximum element can appear at any leaf position, the number of possible values for the index $k$ is 53.

Hence, (A) is the correct option.