Consider a heap containing n elements, where n > 100 and n is odd. Assuming 1-based indexing, which of the following cannot be the index of a leaf node in the heap?

The correct answer is (n−1)/2.

To understand why this is correct, let us analyze the properties of a binary heap represented as an array with 1-based indexing. For any node at index $i$ in a heap containing $n$ elements:
- The left child of the node is located at index $2i$.
- The right child of the node is located at index $2i+1$.
- A node is a leaf node if it has no children, which means its left child index exceeds the total number of elements in the heap: $2i>n$.
- Conversely, a node is an internal (non-leaf) node if it has at least one child, which means:

$2i\le n$

Let us now test the index given in the correct option, which is:

$i=\frac{n-1}{2}$

To determine if this node can be a leaf, we calculate the index of its left child:

$\text{Left Child Index}=2i=2\left(\frac{n-1}{2}\right)=n-1$

Since we are given that $n>100$, it is clear that:

$n-1\le n$

This means the index of the left child, $n-1$, is a valid position within the heap. In fact, since $n$ is odd, the node at index $\frac{n-1}{2}$ also has a right child at index $n$. Because the node at index $\frac{n-1}{2}$ has children, it must be an internal node and cannot be a leaf node in the heap.