Amazon Online Assessment (OA) - Slowest Key
Practice here: https://leetcode.com/problems/slowest-key/
Solution and Explanation
This question asks for the key with largest duration
releaseTimes[i] - releaseTimes[i - 1]. We can simple loop through each release time and calculate its difference to the previous time. If multiple characters have the same duration, we want to use the lexicographically largest.
class Solution: def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str: slowest_key = 'a' longest_duration = 0 n = len(keysPressed) for i in range(n): pressedTime = releaseTimes[i - 1] if i > 0 else 0 duration = releaseTimes[i] - pressedTime if duration == longest_duration: slowest_key = max(slowest_key, keysPressed[i]) elif duration > longest_duration: slowest_key = keysPressed[i] longest_duration = duration return slowest_key