Leetcode simple question: the key with the longest key duration

subject

LeetCode Designed a new keyboard , Testing its availability . The tester will click a series of keys （ A total of n individual ）, One at a time .
Give you a length of n String keysPressed , among keysPressed[i] Represents the... In the test sequence i A pressed key .releaseTimes It's a list in ascending order , among releaseTimes[i] It means that the... Is released i Key time . String and array Subscripts are all from 0 Start . The first 0 Keys at time 0 When pressed , Next, each key just Pressed when the previous key is released .
The tester wants to find the key The longest duration Key . The first i The duration of the second key is releaseTimes[i] - releaseTimes[i - 1] , The first 0 The duration of the second key is releaseTimes[0] .
Be careful , Testing period , The same key can be pressed many times at different times , And the duration of each time may be different .
Please return to the single key The longest duration Key , If there are multiple such keys , Then return to In alphabetical order, the largest The key of .
Example 1：
Input ：releaseTimes = [9,29,49,50], keysPressed = “cbcd”
Output ：“c”
explain ： The key sequence and duration are as follows ：
Press down ‘c’ , The duration of the 9（ Time 0 Press down , Time 9 Release ）
Press down ‘b’ , The duration of the 29 - 9 = 20（ Time to release the last key 9 Press down , Time 29 Release ）
Press down ‘c’ , The duration of the 49 - 29 = 20（ Time to release the last key 29 Press down , Time 49 Release ）
Press down ‘d’ , The duration of the 50 - 49 = 1（ Time to release the last key 49 Press down , Time 50 Release ）
The key with the longest key duration is ‘b’ and ‘c’（ When pressed the second time ）, The duration is 20
‘c’ Arrange in alphabetical order ‘b’ Big , So the answer is ‘c’
Example 2：
Input ：releaseTimes = [12,23,36,46,62], keysPressed = “spuda”
Output ：“a”
explain ： The key sequence and duration are as follows ：
Press down ‘s’ , The duration of the 12
Press down ‘p’ , The duration of the 23 - 12 = 11
Press down ‘u’ , The duration of the 36 - 23 = 13
Press down ‘d’ , The duration of the 46 - 36 = 10
Press down ‘a’ , The duration of the 62 - 46 = 16
The key with the longest key duration is ‘a’ , The duration of the 16
Tips ：
releaseTimes.length == n
keysPressed.length == n
2 <= n <= 1000
1 <= releaseTimes[i] <= 10^9
releaseTimes[i] < releaseTimes[i+1]
keysPressed It's only made up of lowercase letters
source ： Power button （LeetCode）

Their thinking

   Just traverse the array and string at the same time , Calculate the current character key often , Then update the maximum value according to the rules .

class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        releaseTimes.insert(0,0) # For the convenience of calculating the difference, add 0
        MAX=(0,'Z')  # Maximum 
        for i in range(1,len(releaseTimes)):  # Update the maximum value according to the rule 
            if releaseTimes[i]-releaseTimes[i-1]>MAX[0]:
                MAX=(releaseTimes[i]-releaseTimes[i-1],keysPressed[i-1])
            elif releaseTimes[i]-releaseTimes[i-1]==MAX[0] and keysPressed[i-1]>MAX[1]:
                MAX=(MAX[0],keysPressed[i-1])
        return MAX[1]