How do you solve Slowest Key on LeetCode?

Slowest Key (LeetCode #1629) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The key with the longest duration is prioritized, and in case of ties, the lexicographically larger key is chosen.

What is the brute force solution for Slowest Key?

The brute force approach for Slowest Key is Brute Force, with O(n²) time complexity and O(1) space complexity. We can calculate the duration for each keypress and compare them to find the longest duration. This approach is straightforward but inefficient as it checks each key against every other key. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Slowest Key?

Slowest Key uses the following concepts: Array, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Slowest Key?

Always clarify the problem statement and edge cases before jumping into coding. Think about the time complexity and space complexity of your solution. Practice explaining your thought process as you code.

What are common mistakes when solving Slowest Key?

Not considering the first key's duration correctly. Failing to compare keys lexicographically when durations are equal.

#1629

Slowest Key

Easy

ArrayStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can achieve a more efficient solution by calculating the duration in a single pass through the keysPressed and releaseTimes arrays, maintaining the maximum duration and corresponding key as we go.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables for maximum duration and the corresponding key.
2Step 2: Loop through the keysPressed and calculate the duration for each keypress.
3Step 3: Update the maximum duration and key based on the conditions of duration and lexicographical order.

solution.py11 lines

1def slowestKey(releaseTimes, keysPressed):
2    max_duration = 0
3    result_key = ''
4
5    for i in range(len(keysPressed)):
6        duration = releaseTimes[i] if i == 0 else releaseTimes[i] - releaseTimes[i - 1]
7        if duration > max_duration or (duration == max_duration and keysPressed[i] > result_key):
8            max_duration = duration
9            result_key = keysPressed[i]
10
11    return result_key

ℹ

Complexity note: The complexity is O(n) because we only make a single pass through the keysPressed and releaseTimes arrays, calculating durations in constant time.

1The key with the longest duration is prioritized, and in case of ties, the lexicographically larger key is chosen.
2Understanding the relationship between keypress duration and release times is crucial.

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