LeetCode 2933: High-Access Employees Solution

Master LeetCode problem 2933 (High-Access Employees), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2933. High-Access Employees

Medium

Array Hash Table String Sorting

Problem Explanation

Explanation:

We can iterate through each employee's access times and keep track of the count of accesses within each one-hour period.
To do this, we need to convert the access times to minutes for easier comparison.
We can maintain a hashmap where the key is the employee name and value is a list of access times in minutes.
After processing all access times, we iterate through the hashmap to check for employees with three or more accesses within one hour.
Finally, we return the list of high-access employees.

Time Complexity: O(n) where n is the number of access times. Space Complexity: O(n) for the hashmap storing access times.

Solution Code

import java.util.*;

class Solution {
    public List<String> highAccessEmployees(String[][] accessTimes) {
        Map<String, List<Integer>> accessMap = new HashMap<>();

        for (String[] access : accessTimes) {
            String employee = access[0];
            int time = Integer.parseInt(access[1].substring(0, 2)) * 60 + Integer.parseInt(access[1].substring(2));

            accessMap.putIfAbsent(employee, new ArrayList<>());
            accessMap.get(employee).add(time);
        }

        List<String> result = new ArrayList<>();
        for (Map.Entry<String, List<Integer>> entry : accessMap.entrySet()) {
            List<Integer> times = entry.getValue();
            Collections.sort(times);

            for (int i = 0; i < times.size() - 2; i++) {
                if (times.get(i + 2) - times.get(i) <= 60) {
                    result.add(entry.getKey());
                    break;
                }
            }
        }

        return result;
    }
}

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