Here's how we can approach this problem.

Naive Approach

The most straightforward approach is to iterate through the events array, keeping track of the time each button was pressed and calculating the duration for each press. We maintain a dictionary or hash map to store the last time each button was pressed. As we iterate through the events, we update the time taken for each button and keep track of the button with the longest press time.

def longest_button_press_naive(events):
    last_press_time = {}
    max_time = 0
    longest_button = -1

    for index, time in events:
        if index in last_press_time:
            duration = time - last_press_time[index]
        else:
            duration = time

        if duration > max_time:
            max_time = duration
            longest_button = index
        elif duration == max_time and index < longest_button:
            longest_button = index

        last_press_time[index] = time

    return longest_button

Complexity Analysis

• Time Complexity: O(n), where n is the number of events, as we iterate through the events array once.
• Space Complexity: O(k), where k is the number of unique buttons pressed. In the worst case, every event corresponds to a unique button, making it O(n).

Optimal Approach

Since the naive approach already has a linear time complexity, it is already quite efficient. We can refine the code for clarity and conciseness, but the fundamental algorithm remains the same.

def longest_button_press_optimal(events):
    button_times = {}
    max_duration = 0
    longest_button = -1

    for index, time in events:
        duration = time - button_times.get(index, 0)
        if index not in button_times:
            duration = time

        if duration > max_duration:
            max_duration = duration
            longest_button = index
        elif duration == max_duration and index < longest_button:
            longest_button = index

        button_times[index] = time

    return longest_button

Complexity Analysis

• Time Complexity: O(n), where n is the number of events.
• Space Complexity: O(k), where k is the number of unique buttons. In the worst case, every event corresponds to a unique button, making it O(n).

Edge Cases

• Empty Input: If the input events array is empty, the problem statement doesn't define what to return. We might return -1 or raise an exception based on the requirements.
• Single Event: If there is only one event, the duration is simply the time at which the button was pressed. The algorithm should handle this correctly.
• Same Longest Time: The problem specifies that if multiple buttons have the same longest time, we should return the button with the smallest index. The algorithm incorporates this condition using elif duration == max_time and index < longest_button:.

Summary

Both the naive and optimal approaches involve iterating through the events array, keeping track of the last press time for each button, and updating the maximum duration and corresponding button index. The time complexity is O(n), and the space complexity is O(k), where k is the number of unique buttons.