leetcode-1604. Alert Using Same Key-Card Three or More Times in a One Hour Period

题目

LeetCode company workers use key-cards to unlock office doors. Each time a worker uses their key-card, the security system saves the worker’s name and the time when it was used. The system emits an alert if any worker uses the key-card three or more times in a one-hour period.

You are given a list of strings keyName and keyTime where [keyName[i], keyTime[i]] corresponds to a person’s name and the time when their key-card was used in a single day.

Access times are given in the 24-hour time format “HH:MM”, such as “23:51” and “09:49”.

Return a list of unique worker names who received an alert for frequent keycard use. Sort the names in ascending order alphabetically.

Notice that “10:00” - “11:00” is considered to be within a one-hour period, while “22:51” - “23:52” is not considered to be within a one-hour period.

Example 1:

Input: keyName = ["daniel","daniel","daniel","luis","luis","luis","luis"], keyTime = ["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]
Output: ["daniel"]
Explanation: "daniel" used the keycard 3 times in a one-hour period ("10:00","10:40", "11:00").

Example 2:

Input: keyName = ["alice","alice","alice","bob","bob","bob","bob"], keyTime = ["12:01","12:00","18:00","21:00","21:20","21:30","23:00"]
Output: ["bob"]
Explanation: "bob" used the keycard 3 times in a one-hour period ("21:00","21:20", "21:30").

Constraints:

1 <= keyName.length, keyTime.length <= 105
keyName.length == keyTime.length
keyTime[i] is in the format "HH:MM".
[keyName[i], keyTime[i]] is unique.
1 <= keyName[i].length <= 10
keyName[i] contains only lowercase English letters.

解题思路

纯粹的模拟题，把每个员工的时间存在字典里，然后排序，再比较字符串即可。对每个时间计算后面的2个时间是否在它的+1小时内

注意字符串的填充方式，需要对个位数的小时左边填充0，这样才比较好比较

看了官方题解，把小时转化为分钟也是一个比较好的解法

时间复杂度$o(nlogn)$（主要在排序上）
空间复杂度$o(n)$

代码

class Solution:
    def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
        employer_info = {}
        for name, time in zip(keyName, keyTime):
            if name not in employer_info:
                employer_info[name] = []
            employer_info[name].append(time)
        res = set()
        # sort time
        for name, info in employer_info.items():
            info.sort()
            for index, each_time in enumerate(info[:-2]):
                longest_time = f'{1 + int(each_time[:2]):0>2}{each_time[2:]}'
                if longest_time >= info[index + 2]:
                    res.add(name)
        return list(sorted(res))