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.
1class Solution:
2 def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
3 slowest_key = 'a'
4 longest_duration = 0
5 n = len(keysPressed)
6
7 for i in range(n):
8 pressedTime = releaseTimes[i - 1] if i > 0 else 0
9 duration = releaseTimes[i] - pressedTime
10 if duration == longest_duration:
11 slowest_key = max(slowest_key, keysPressed[i])
12 elif duration > longest_duration:
13 slowest_key = keysPressed[i]
14 longest_duration = duration
15
16 return slowest_key
171class Solution {
2 public char slowestKey(int[] releaseTimes, String keysPressed) {
3 int n = releaseTimes.length;
4 char[] keys = keysPressed.toCharArray();
5 char ans = 'a';
6 int longestDuration = 0;
7 int pressedTime;
8 int duration;
9 for (int i = 0; i < n; i++) {
10 if (i == 0) {
11 pressedTime = 0;
12 } else {
13 pressedTime = releaseTimes[i - 1];
14 }
15 duration = releaseTimes[i] - pressedTime;
16 if (duration == longestDuration) {
17 if (ans < keys[i]) ans = keys[i];
18 } else if (duration > longestDuration) {
19 ans = keys[i];
20 longestDuration = duration;
21 }
22 }
23 return ans;
24 }
25}
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