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Formatted question description: https://leetcode.ca/all/1604.html

# 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/
// 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;
}
};


Java

• class Solution {
public List<String> alertNames(String[] keyName, String[] keyTime) {
Map<String, List<String>> map = new HashMap<String, List<String>>();
int length = keyName.length;
for (int i = 0; i < length; i++) {
String name = keyName[i];
String time = keyTime[i];
List<String> list = map.getOrDefault(name, new ArrayList<String>());
map.put(name, list);
}
Set<String> keySet = map.keySet();
for (String name : keySet) {
List<String> list = map.get(name);
Collections.sort(list);
int size = list.size();
for (int i = 2; i < size; i++) {
String time1 = list.get(i - 2), time2 = list.get(i);
int hour1 = Integer.parseInt(time1.substring(0, 2)), minute1 = Integer.parseInt(time1.substring(3));
int hour2 = Integer.parseInt(time2.substring(0, 2)), minute2 = Integer.parseInt(time2.substring(3));
int difference = (hour2 * 60 + minute2) - (hour1 * 60 + minute1);
if (difference <= 60) {
break;
}
}
}
}
}

• // OJ: https://leetcode.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period/
// 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;
}
};

• class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:
d = defaultdict(list)
for name, t in zip(keyName, keyTime):
t = int(t[:2]) * 60 + int(t[3:])
d[name].append(t)
ans = []
for name, ts in d.items():
if (n := len(ts)) > 2:
ts.sort()
for i in range(n - 2):
if ts[i + 2] - ts[i] <= 60:
ans.append(name)
break
ans.sort()
return ans

############

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

class Solution:
def alertNames(self, keyName: List[str], keyTime: List[str]) -> List[str]:

def convertTime(time):
return int("".join(time.split(":")))

dic = {}

for name,time in zip(keyName,keyTime):
time = convertTime(time)

if name not in dic:
dic[name] = [time]
else:
dic[name].append(time)

res = []

for name in dic:
timeList = sorted(dic[name])
tmp = []
c = 0

for time in timeList:

while tmp and time - tmp > 100:
tmp.pop(0)
if c > 0:
c -= 1

c += 1
tmp.append(time)

if c >= 3:
res.append(name)
break

return sorted(res)