How do you solve Button with Longest Push Time on LeetCode?

Button with Longest Push Time (LeetCode #3386) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each button's time is calculated based on its last press time.

What is the time complexity of Button with Longest Push Time?

The optimal solution for Button with Longest Push Time runs in O(n) time and uses O(n) space. This complexity is linear because we only traverse the events list once, storing the last press time for each button.

What is the brute force solution for Button with Longest Push Time?

The brute force approach for Button with Longest Push Time is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we calculate the time taken for each button press by comparing it with every other button press. This is straightforward but inefficient as it checks all possible pairs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Button with Longest Push Time?

Button with Longest Push Time uses the following concepts: Array. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Button with Longest Push Time?

Clarify the problem statement and constraints before coding. Think about edge cases, such as buttons pressed only once. Explain your thought process clearly while coding.

What are common mistakes when solving Button with Longest Push Time?

Not updating the last press time correctly. Confusing the time taken with the total time of all presses.

#3386

Button with Longest Push Time

Easy

ArrayHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

In the optimal solution, we traverse the events list once, calculating the time taken for each button press in a single pass. This is efficient and avoids unnecessary comparisons.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dictionary to store the last press time for each button.
2Step 2: Loop through the events and calculate the time taken for each button press.
3Step 3: Update the maximum time and corresponding button index if the current time taken is greater.

solution.py16 lines

1def longest_push_time(events):
2    button_times = {}
3    max_time = 0
4    result_index = -1
5
6    for index, time in events:
7        if index not in button_times:
8            button_times[index] = time
9        else:
10            time_taken = time - button_times[index]
11            button_times[index] = time
12            if time_taken > max_time or (time_taken == max_time and index < result_index):
13                max_time = time_taken
14                result_index = index
15
16    return result_index

ℹ

Complexity note: This complexity is linear because we only traverse the events list once, storing the last press time for each button.

1Each button's time is calculated based on its last press time.
2We need to track both the time taken and the button index efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.