LeetCode-in-Java
2933. High-Access Employees
Medium
You are given a 2D 0-indexed array of strings, access_times, with size n. For each i where 0 <= i <= n - 1, access_times[i][0] represents the name of an employee, and access_times[i][1] represents the access time of that employee. All entries in access_times are within the same day.
The access time is represented as four digits using a 24-hour time format, for example, "0800" or "2250".
An employee is said to be high-access if he has accessed the system three or more times within a one-hour period.
Times with exactly one hour of difference are not considered part of the same one-hour period. For example, "0815" and "0915" are not part of the same one-hour period.
Access times at the start and end of the day are not counted within the same one-hour period. For example, "0005" and "2350" are not part of the same one-hour period.
Return a list that contains the names of high-access employees with any order you want.
Example 1:
Input: access_times = [[“a”,”0549”],[“b”,”0457”],[“a”,”0532”],[“a”,”0621”],[“b”,”0540”]]
Output: [“a”]
Explanation: “a” has three access times in the one-hour period of [05:32, 06:31] which are 05:32, 05:49, and 06:21. But “b” does not have more than two access times at all. So the answer is [“a”].
Example 2:
Input: access_times = [[“d”,”0002”],[“c”,”0808”],[“c”,”0829”],[“e”,”0215”],[“d”,”1508”],[“d”,”1444”],[“d”,”1410”],[“c”,”0809”]]
Output: [“c”,”d”]
Explanation: “c” has three access times in the one-hour period of [08:08, 09:07] which are 08:08, 08:09, and 08:29. “d” has also three access times in the one-hour period of [14:10, 15:09] which are 14:10, 14:44, and 15:08. However, “e” has just one access time, so it can not be in the answer and the final answer is [“c”,”d”].
Example 3:
Input: access_times = [[“cd”,”1025”],[“ab”,”1025”],[“cd”,”1046”],[“cd”,”1055”],[“ab”,”1124”],[“ab”,”1120”]]
Output: [“ab”,”cd”]
Explanation: “ab” has three access times in the one-hour period of [10:25, 11:24] which are 10:25, 11:20, and 11:24. “cd” has also three access times in the one-hour period of [10:25, 11:24] which are 10:25, 10:46, and 10:55. So the answer is [“ab”,”cd”].
Constraints:
1 <= access_times.length <= 100access_times[i].length == 21 <= access_times[i][0].length <= 10access_times[i][0]consists only of English small letters.access_times[i][1].length == 4access_times[i][1]is in 24-hour time format.access_times[i][1]consists only of'0'to'9'.
Solution
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class Solution {
private boolean isPossible(int a, int b) {
int hb = b / 100;
int ha = a / 100;
int mind = b % 100;
int mina = a % 100;
if (hb == 23 && ha == 0) {
return false;
}
if (hb - ha > 1) {
return false;
}
if (hb - ha == 1) {
mind += 60;
}
return mind - mina < 60;
}
private boolean isHighAccess(List<Integer> list) {
if (list.size() < 3) {
return false;
}
int i = 0;
int j = 1;
int k = 2;
while (k < list.size()) {
int a = list.get(i++);
int b = list.get(j++);
int c = list.get(k++);
if (isPossible(a, c) && isPossible(b, c) && isPossible(a, b)) {
return true;
}
}
return false;
}
private int stringToInt(String str) {
int i = 1000;
int val = 0;
for (char ch : str.toCharArray()) {
int n = ch - '0';
val += i * n;
i = i / 10;
}
return val;
}
public List<String> findHighAccessEmployees(List<List<String>> accessTimes) {
HashMap<String, List<Integer>> map = new HashMap<>();
for (List<String> list : accessTimes) {
List<Integer> temp = map.getOrDefault(list.get(0), new ArrayList<>());
int val = stringToInt(list.get(1));
temp.add(val);
map.put(list.get(0), temp);
}
List<String> ans = new ArrayList<>();
for (Map.Entry<String, List<Integer>> entry : map.entrySet()) {
List<Integer> temp = entry.getValue();
Collections.sort(temp);
if (isHighAccess(temp)) {
ans.add(entry.getKey());
}
}
return ans;
}
}