1629.SlowestKey.py

"""
    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:
    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:
    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] <= 10**9
        - releaseTimes[i] < releaseTimes[i+1]
        - keysPressed contains only lowercase English letters.
"""
#Difficulty: Easy
#101 / 101 test cases passed.
#Runtime: 44 ms
#Memory Usage: 14.4 MB

#Runtime: 44 ms, faster than 99.79% of Python3 online submissions for Slowest Key.
#Memory Usage: 14.4 MB, less than 6.39% of Python3 online submissions for Slowest Key.

class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        key = None
        long_press = 0
        prev_time = 0
        for time, pressed_key in zip(releaseTimes, keysPressed):
            current_press = time - prev_time
            prev_time = time
            if current_press >= long_press:
                key = pressed_key
                long_press = current_press
        return key