# [1629. Slowest Key](https://leetcode.com/problems/slowest-key/)
###### tags: `Array`, `String`, `Amazon OA`

Description:
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.*

Solution:

class Solution {
    public char slowestKey(int[] releaseTimes, String keysPressed) {
        char[] arr = keysPressed.toCharArray();
        int max = releaseTimes[0];
        char res = arr[0];
        for (int i = 1; i < arr.length; i++) {
            int dur = releaseTimes[i] - releaseTimes[i - 1];
            if (max == dur && arr[i] > res) {
                res = arr[i];
            } else if (max < dur) {
                res = arr[i];
                max = dur;
            }
        }
        return res;
    }
}