1604

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

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 "23:51" - "00:10" 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").

Example 3:

Input: keyName = ["john","john","john"], keyTime = ["23:58","23:59","00:01"]
Output: []

Example 4:

Input: keyName = ["leslie","leslie","leslie","clare","clare","clare","clare"], keyTime = ["13:00","13:20","14:00","18:00","18:51","19:30","19:49"]
Output: ["clare","leslie"]

 

Constraints:

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

Related Topics:
String, Ordered Map

Solution 1. Hash Map

Use unordered_map<string, vector<int>> to group the times by name.

For the times of each person, sort the times. If there is any times[i] - times[i - 2] <= 60, then the person should get alerted.

// OJ: https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<string> alertNames(vector<string>& keyName, vector<string>& keyTime) {
        unordered_map<string, vector<int>> m;
        for (int i = 0; i < keyName.size(); ++i) {
            auto &key = keyName[i], &time = keyTime[i];
            int t = stoi(time.substr(0, 2)) * 60 + stoi(time.substr(3));
            m[key].push_back(t);
        }
        vector<string> ans;
        for (auto &[key, times] : m) {
            sort(begin(times), end(times));
            for (int i = 2; i < times.size(); ++i) {
                if (times[i] - times[i - 2] > 60) continue;
                ans.push_back(key);
                break;
            }
        }
        sort(begin(ans), end(ans));
        return ans;
    }
};