An array A of length n with distinct elements is said to be bitonic if there is an index 1 = i = n  such that A[1..i]A[1..i] is sorted in the non-decreasing order and A[i+1..n] is sorted in the non-increasing order. Which ONE of the following represents the best possible asymptotic bound for the worst-case number of comparisons by an algorithm that searches for an element in a bitonic array A?

The correct option is Θ(log n).

Let us understand why the worst-case number of comparisons for searching an element in a bitonic array is bounded by $\Theta (\log n)$.

A bitonic array is an array that consists of distinct elements, which first increases monotonically up to a certain maximum element (the peak element at index $i$), and then decreases monotonically. Formally, $A[1..i]$ is sorted in ascending order and $A[i+1..n]$ is sorted in descending order.

To search for a target element in such an array efficiently, we can break down the process into two main phases, both utilizing modified binary search:

Step 1: Finding the peak element (the bitonic point)
We can find the index $i$ of the peak element using a modified binary search. At each step, we look at the middle element $A\left[m\right]$ and compare it with its neighbors:
- If $A\left[m\right]>A[m-1]$ and $A\left[m\right]>A[m+1]$, then $A\left[m\right]$ is the peak.
- If $A\left[m\right]>A[m-1]$ and $A\left[m\right]<A[m+1]$, the peak must lie in the right half: $A[m+1..n]$.
- If $A\left[m\right]<A[m-1]$ and $A\left[m\right]>A[m+1]$, the peak must lie in the left half: $A[1..m-1]$.
Since we halve the search space at each step, finding the peak element takes $\Theta (\log n)$ comparisons.

Step 2: Searching for the target element
Once the peak index $i$ is identified, the array is split into two sorted subarrays:
1. An ascending sorted subarray $A[1..i]$.
2. A descending sorted subarray $A[i+1..n]$.
We perform a standard binary search for the target element on the ascending subarray, which takes $\Theta (\log i)$ comparisons. If not found, we perform a binary search (adapted for descending order) on the descending subarray, which takes $\Theta (\log (n-i\left)\right)$ comparisons.

Total Time Complexity:
The overall worst-case comparisons is the sum of the comparisons in both steps:
$T\left(n\right)=T_{\text{find\_peak}}+T_{\text{search\_ascending}}+T_{\text{search\_descending}}$
$T\left(n\right)=\Theta (\log n)+\Theta (\log i)+\Theta (\log (n-i\left)\right)$
Since $i\le n$ and $n-i\le n$, the complexity simplifies to:
$T\left(n\right)=\Theta (\log n)$

Thus, the best possible asymptotic bound for the worst-case number of comparisons is $\Theta (\log n)$.