Button with Longest Push Time Interview Question for Amazon

You are given a 2D array events which represents a sequence of events where a child pushes a series of buttons on a keyboard. Each events[i] = [index_i, time_i] indicates that the button at index index_i was pressed at time time_i. The array is sorted in increasing order of time. The time taken to press a button is the difference in time between consecutive button presses. The time for the first button is simply the time at which it was pressed. Return the index of the button that took the longest time to push. If multiple buttons have the same longest time, return the button with the smallest index.

For example:

events = [[1,2],[2,5],[3,9],[1,15]]

In the example above, the output is 1.

Here's the breakdown:

Button with index 1 is pressed at time 2.
Button with index 2 is pressed at time 5, so it took 5 - 2 = 3 units of time.
Button with index 3 is pressed at time 9, so it took 9 - 5 = 4 units of time.
Button with index 1 is pressed again at time 15, so it took 15 - 9 = 6 units of time.

As another example:

events = [[10,5],[1,7]]

In the example above, the output is 10.

Here's the breakdown:

Button with index 10 is pressed at time 5.
Button with index 1 is pressed at time 7, so it took 7 - 5 = 2 units of time.

## Optimal Solution

This problem can be solved efficiently by iterating through the `events` array once and keeping track of the maximum time difference encountered so far, along with the corresponding index. Here's the algorithm:

1.  Initialize `max_time` to 0 and `result_index` to -1 (or any invalid index).
2.  Iterate through the `events` array from the second element.
3.  For each element `events[i]`, calculate the time difference between `events[i][1]` and `events[i-1][1]`.
4.  If the time difference is greater than `max_time`, update `max_time` with the new time difference and `result_index` with `events[i][0]`. Also, if the time difference is the same as max_time, check which index is smaller and update accordingly.
5.  For the first element, we assume the time taken to press the button is the time at which it was pressed, so we initialize `max_time` with `events[0][1]` and `result_index` with `events[0][0]`. This handles edge cases where the longest time is the first button.
6.  Return `result_index`.

### Code

```python
def longest_time_to_push(events):
    max_time = events[0][1]
    result_index = events[0][0]

    for i in range(1, len(events)):
        time_diff = events[i][1] - events[i-1][1]
        if time_diff > max_time:
            max_time = time_diff
            result_index = events[i][0]
        elif time_diff == max_time and events[i][0] < result_index:
            result_index = events[i][0]

    return result_index

Example Usage:

events1 = [[1,2],[2,5],[3,9],[1,15]]
events2 = [[10,5],[1,7]]

print(longest_time_to_push(events1)) # Output: 1
print(longest_time_to_push(events2)) # Output: 10

Naive Solution

A naive solution would involve calculating the time taken for each button press and storing them in a dictionary or a separate list. Then, iterate over the calculated times to find the maximum time and its corresponding index.

Create a dictionary to store the total time taken for each button.
Iterate through the events array.
For the first event, store the time in the dictionary.
For subsequent events, calculate the time difference between the current event and the previous event and add it to the corresponding button's total time.
Iterate through the dictionary to find the button with the maximum time.
Return the index of the button with the maximum time.

This approach is less efficient due to the overhead of using a dictionary and requires additional iterations.

Code for Naive Solution

def longest_time_to_push_naive(events):
    button_times = {}
    
    for i in range(len(events)):
        index, time = events[i]
        if index not in button_times:
            button_times[index] = 0
            
        if i == 0:
            button_times[index] = time
        else:
            button_times[index] += time - events[i-1][1]
    
    max_time = 0
    result_index = -1
    
    for index, time in button_times.items():
        if time > max_time:
            max_time = time
            result_index = index
        elif time == max_time and index < result_index: #Handling the situation where both have same time, and we have to choose smallest index
          result_index = index
            
    return result_index

Big(O) Time Complexity Analysis

Optimal Solution: The algorithm iterates through the events array once, so the time complexity is O(n), where n is the number of events.
Naive Solution: The naive solution iterates through the events array once to calculate the time for each button and then iterates through the dictionary, resulting in a time complexity of O(n + k), where n is the number of events and k is the number of unique buttons. In the worst case, k can be equal to n, so the time complexity could be O(2n), which simplifies to O(n).

Big(O) Space Complexity Analysis

Optimal Solution: The algorithm uses a constant amount of extra space to store max_time and result_index, so the space complexity is O(1).
Naive Solution: The naive solution uses a dictionary to store the total time for each button. In the worst case, where all buttons are unique, the space complexity is O(n), where n is the number of events.

Edge Cases and Handling

Empty Input: If the input events array is empty, the algorithm should return a default value or raise an exception, depending on the specific requirements.

def longest_time_to_push(events):
    if not events:
        return -1  # Or raise an exception

Single Event: If the input events array contains only one event, the algorithm should return the index of that event.

def longest_time_to_push(events):
    if len(events) == 1:
        return events[0][0]

Same Time: If two consecutive events have the same time, the time difference will be zero. The algorithm should handle this case correctly.
Large Time Values: The algorithm should be able to handle large time values without overflowing. Python handles large integers well, but in other languages, it may be necessary to use appropriate data types.
Duplicate indices: The algorithm must be able to handle duplicate indices correctly to determine which index took the longest.