[leetcode #1629] Slowest Key

Problem

A newly designed keypad was tested, where a tester pressed a sequence of n keys, one at a time.

You are given a string keysPressed of length n, where keysPressed[i] was the ith key pressed in the testing sequence, and a sorted list releaseTimes, where releaseTimes[i] was the time the ith key was released. Both arrays are 0-indexed. The 0th key was pressed at the time 0, and every subsequent key was pressed at the exact time the previous key was released.

The tester wants to know the key of the keypress that had the longest duration. The ith keypress had a duration of releaseTimes[i] - releaseTimes[i - 1], and the 0th keypress had a duration of releaseTimes[0].

Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key may not have had the same duration.

Return the key of the keypress that had the longest duration. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.

Example 1:

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output: "c"
Explanation: The keypresses were as follows:
Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
'c' is lexicographically larger than 'b', so the answer is 'c'.

Example 2:

Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
Output: "a"
Explanation: The keypresses were as follows:
Keypress for 's' had a duration of 12.
Keypress for 'p' had a duration of 23 - 12 = 11.
Keypress for 'u' had a duration of 36 - 23 = 13.
Keypress for 'd' had a duration of 46 - 36 = 10.
Keypress for 'a' had a duration of 62 - 46 = 16.
The longest of these was the keypress for 'a' with duration 16.

Constraints:

・ releaseTimes.length == n
・ keysPressed.length == n
・ 2 <= n <= 1000
・ 1 <= releaseTimes[i] <= 10⁹
・ releaseTimes[i] < releaseTimes[i+1]
・ keysPressed contains only lowercase English letters.

Idea

알고리즘 문제가 아니고 영어 독해 문제가 아닐까. 쉬운 문제인데 대충 읽고 풀려다 두 번정도 틀려서 당황했다. 처음에는 Release time이 같은 경우 사전정렬을 했을 때 뒤에 오는 알파벳을 리턴해야 하는 걸 놓쳤고, 두 번째는 release time의 합이 아니라 가장 큰 release time을 가진 알파벳을 리턴하는 걸 몰랐다.

쉬운 문제라고 얕보지 말고 심사숙고한 뒤 문제를 푸니 만족할만한 결과가 나왔다.

알고리즘은 다음과 같다.

1. releaseTimes와 keysPressed를 탐색한다.
   a. 첫 번째 탐색일 경우 maxChar를 keysPressed의 0번째 index 값으로, maxDuration을 releaseTimes[0]으로 설정한다.
   b. 이후에는 releaseTime을 releaseTimes[i] - releaseTimes[i-1]로 설정한다. releaseTime이 maxDuration보다 크거나, maxDuration과 같고 i번째 문자가 maxChar보다 클 경우 maxChar와 maxDuration을 바꿔준다.
2. maxChar를 리턴한다.

Solution

class Solution {
    public char slowestKey(int[] releaseTimes, String keysPressed) {
        char maxChar = 'a';
        int maxDuration = 0;

        for (int i=0; i < releaseTimes.length; i++) {
            if (i == 0) {
                maxChar = keysPressed.charAt(i); 
                maxDuration = releaseTimes[i];
            }
            else {
                int releaseTime = releaseTimes[i] - releaseTimes[i-1];
                if (maxDuration < releaseTime || (maxDuration == releaseTime && maxChar < keysPressed.charAt(i))) {
                    maxChar = keysPressed.charAt(i); 
                    maxDuration = releaseTimes[i] - releaseTimes[i-1];                    
                }
            }
        }

        return maxChar;
    }
}

Reference

https://leetcode.com/problems/slowest-key/