Button with Longest Push Time

Easy

Asked by:

29 views

Topics:

Arrays

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.

Example 1:

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

Output: 1

Explanation:

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.

Example 2:

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

Output: 10

Explanation:

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.

Constraints:

1 <= events.length <= 1000
events[i] == [index_i, time_i]
1 <= index_i, time_i <= 10⁵
The input is generated such that events is sorted in increasing order of time_i.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected data type and format of the input? Specifically, what data types are the `timestamps` and `button_presses`?
Can the `timestamps` array be empty or contain negative values?
Is it possible for multiple buttons to have the same longest push time? If so, which button should I return?
Is the `timestamps` array guaranteed to be sorted in ascending order?
Is there a maximum number of button presses, and if so, what is it?

Brute Force Solution

Approach

The brute force approach means we're going to try every single possible combination. We'll go through each button press individually and calculate how long that button was pressed for by looking at the time it was released. Then, we'll compare all the durations to find the biggest one.

Here's how the algorithm would work step-by-step:

Look at the first button press.
Calculate how long that button was pressed by subtracting its press time from its release time.
Store that duration somewhere.
Now, look at the next button press.
Calculate the duration of that press in the same way.
Compare the new duration to the duration we stored earlier.
If the new duration is longer, replace the stored duration with the new one.
Keep doing this for every single button press.
After checking all the button presses, the duration we have stored will be the longest press time.

Code Implementation

def find_longest_push_time(presses, releases):
    longest_duration_so_far = 0

    for index in range(len(presses)): 
        # Iterate over all button presses to find the longest duration

        button_press_time = presses[index]
        button_release_time = releases[index]

        button_duration = button_release_time - button_press_time
        # Calculate the duration of the current button press

        if button_duration > longest_duration_so_far:
            # Check if current press is the longest so far

            longest_duration_so_far = button_duration

    return longest_duration_so_far

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through each button press in the input array once to calculate its duration and compare it with the current maximum duration. The input size, n, represents the number of button presses. Since we perform a fixed number of operations (subtraction, comparison) for each button press, the time complexity is directly proportional to the number of button presses. Therefore, the total number of operations is approximately n, and the time complexity simplifies to O(n).

Space Complexity

O(1) – The provided solution only stores a single duration value to keep track of the longest press time encountered so far. No additional data structures like arrays, hash maps, or lists are used to store intermediate results or visited elements. Therefore, the auxiliary space used remains constant regardless of the number of button presses (N), leading to a space complexity of O(1).

Optimal Solution

Approach

The problem asks us to figure out which button was held down for the longest amount of time. The key is to go through the provided records one by one and keep track of the time each button was pressed.

Here's how the algorithm would work step-by-step:

Look at the starting time and button pressed in the first record.
Go to the next record and find out when that button was pressed.
Calculate how long the first button was held down by subtracting the first button's start time from the time the next button was pressed.
Compare the calculated time with the longest time we've seen so far for any button. If it's longer, remember this new button and its time.
Repeat steps 2-4 for each record, always calculating the duration and comparing it to the longest time we've recorded. Move through all of the records until you hit the last entry.
For the last button in the records, subtract the current timestamp from the time the last button was pressed. Compare the result to the current maximum to check if the last button was held down the longest.
After checking all records, the button and duration we remember is the answer.

Code Implementation

def button_with_longest_push(records):
    longest_time = 0
    longest_button = None
    start_time = records[0][0]
    start_button = records[0][1]

    # Iterate through the records to find the longest push time.
    for i in range(1, len(records)):
        current_time = records[i][0]
        current_button = records[i][1]

        time_difference = current_time - start_time

        # Compare current button's time with the longest so far.
        if time_difference > longest_time:
            longest_time = time_difference
            longest_button = start_button

        start_time = current_time
        start_button = current_button

    # Handle the last button press to calculate duration until the end
    time_difference = records[-1][0] - start_time

    # Check if the last button had the longest duration
    if time_difference > longest_time:
        longest_time = time_difference
        longest_button = start_button

    return longest_button

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input records once. In each iteration, it performs constant-time operations such as calculating the duration a button was pressed and comparing it to the current maximum. Since the number of operations is directly proportional to the number of records (n), the time complexity is O(n).

Space Complexity

O(1) – The provided solution primarily uses a few variables to keep track of the longest duration seen so far and the corresponding button. It does not create any auxiliary data structures like arrays, lists, or hash maps whose size depends on the number of records (N). Regardless of how many records there are, the algorithm only needs constant extra space. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty releaseTimes array	Return -1 since there are no button presses to analyze.
Null or empty pressedKeys string	Return null character since there are no button presses to analyze.
releaseTimes array and pressedKeys string have different lengths	Return null or throw an IllegalArgumentException indicating the input is invalid since we need the same number of button presses as times.
releaseTimes array contains non-positive values (zero or negative)	Return null or throw IllegalArgumentException, as release times should be monotonically increasing and positive.
releaseTimes array is not monotonically increasing (e.g., [1, 3, 2])	Return null or throw IllegalArgumentException because the release times should be ordered.
Maximum array size causes integer overflow when calculating differences	Use long to store the differences to prevent integer overflow during calculation.
Multiple characters have the same longest push time	Choose the lexicographically largest character as the problem specifies to resolve ties.
Single button push	Return the character corresponding to that single push, as it is the only possible answer.

Null or empty releaseTimes array

How to Handle:

Return -1 since there are no button presses to analyze.

Null or empty pressedKeys string

How to Handle:

Return null character since there are no button presses to analyze.

releaseTimes array and pressedKeys string have different lengths

How to Handle:

Return null or throw an IllegalArgumentException indicating the input is invalid since we need the same number of button presses as times.

releaseTimes array contains non-positive values (zero or negative)

How to Handle:

Return null or throw IllegalArgumentException, as release times should be monotonically increasing and positive.

releaseTimes array is not monotonically increasing (e.g., [1, 3, 2])

How to Handle:

Return null or throw IllegalArgumentException because the release times should be ordered.

Maximum array size causes integer overflow when calculating differences

How to Handle:

Use long to store the differences to prevent integer overflow during calculation.

Multiple characters have the same longest push time

How to Handle:

Choose the lexicographically largest character as the problem specifies to resolve ties.

Single button push

How to Handle:

Return the character corresponding to that single push, as it is the only possible answer.