Intuition:

To find the maximum element in a bitonic array, we can use a modified binary search algorithm. Since the array is bitonic, we know that there exists a peak element where the increasing sequence ends and the decreasing sequence begins. We can perform binary search to find this peak element, as it marks the maximum element in the array. Once we find the peak element, we can return it as the maximum element in the bitonic array.

Solution:

from typing import List

def bitonic_array_maximum(arr: List[int]) -> int:
    # Perform binary search to find the peak element
    left, right = 0, len(arr) - 1
    while left < right:
        mid = (left + right) // 2
        if arr[mid] < arr[mid + 1]:
            # Move towards the increasing sequence
            left = mid + 1
        else:
            # Move towards the decreasing sequence
            right = mid
    # The peak element marks the maximum element in the bitonic array
    return arr[left]

# Test case
arr = [1, 3, 8, 12, 4, 2]
result = bitonic_array_maximum(arr)
print(result)  # Output should be 12

Time Complexity:

Binary search is performed to find the peak element, and each iteration divides the search range in half. The time complexity of this approach is O(log(n)), where n is the number of elements in the input array.

Space Complexity:

The space complexity is O(1) as we use only a constant amount of additional space for variables.

Solution 2: Brute Force

A brute-force approach to finding the maximum element in a bitonic array would involve scanning through the entire array to identify the point where the strictly increasing sequence transitions into the strictly decreasing sequence. Once we find this transition point, the maximum element in the array would be the element immediately preceding it. This approach would iterate over each element in the array, comparing it with the next element to identify the transition point.

Solution:

from typing import List

def bitonic_array_maximum(arr: List[int]) -> int:
    # Initialize the maximum element
    max_element = arr[0]
    
    # Iterate through the array to find the transition point
    for i in range(1, len(arr)):
        # If the current element is greater than the maximum element so far
        # update the maximum element
        if arr[i] > max_element:
            max_element = arr[i]
        # If the current element is less than or equal to the previous element
        # we have found the transition point
        elif arr[i] <= arr[i - 1]:
            break
    
    # Return the maximum element
    return max_element

# Test case
arr = [1, 3, 8, 12, 4, 2]
result = bitonic_array_maximum(arr)
print(result)  # Output should be 12

Time Complexity:

We iterate through the array once to identify the transition point. The time complexity of this brute-force approach is O(n), where n is the number of elements in the input array.

Space Complexity:

The space complexity is O(1) as we use only a constant amount of additional space for variables.