What is the time complexity of High-Access Employees?

The optimal solution has a time complexity of O(nlogn).

What data structures are used to solve High-Access Employees?

This problem uses Array, Hash Table, String, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving High-Access Employees?

Insight 1: The problem requires grouping access times by employee and then checking for overlapping intervals (within a one-hour range). Insight 2: Sorting the access times for each employee is crucial to efficiently check for the one-hour period constraint.

Which companies ask LeetCode 2933. High-Access Employees in interviews?

This problem is frequently asked at Atlassian.

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High-Access Employees - LeetCode 2933 Solution

Q: How to solve High-Access Employees?

The solution approach involves the following steps: First, group the access times by employee using a hash map (dictionary). Then, for each employee, sort their access times numerically. Finally, iterate through the sorted access times and check if there are at least three access times within a one-hour (100 minutes in the time representation) window. If an employee meets this condition, add their name to the list of high-access employees.

High-Access Employees - Complete Solution Guide

High-Access Employees is LeetCode problem 2933, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem requires identifying 'high-access' employees from a list of employee access times. An employee is considered high-access if they have accessed the system three or more times within a one-hour period. The access times are given in a 24-hour format string (e.g., '0800', '2250'). The input is a 2D array of strings, where each row represents an employee's name and access time. The output should be a list of the names of high-access employees, with no duplicates and in any order. Crucially, times must fall within the same one-hour window to be counted.

Solution Approach

The solution approach involves the following steps: First, group the access times by employee using a hash map (dictionary). Then, for each employee, sort their access times numerically. Finally, iterate through the sorted access times and check if there are at least three access times within a one-hour (100 minutes in the time representation) window. If an employee meets this condition, add their name to the list of high-access employees.

Step-by-Step Algorithm

Step 1: Create a hash map (dictionary) to store access times grouped by employee name.
Step 2: Iterate through the input `access_times` array and populate the hash map. Convert the access time strings to integers and store them in a list associated with the employee's name.
Step 3: Create an empty list to store the names of high-access employees.
Step 4: Iterate through the hash map, processing each employee and their access times.
Step 5: Sort the access times for each employee in ascending order.
Step 6: Check if the employee has at least three access times. If not, skip to the next employee.
Step 7: Iterate through the sorted access times using a sliding window of size 3. For each window, check if the difference between the earliest and latest access time is less than 100 (one hour).
Step 8: If the difference is less than 100, it means there are three access times within one hour. Add the employee's name to the list of high-access employees and break out of the inner loop (as the employee is confirmed to be high-access).
Step 9: Return the list of high-access employees.

Key Insights

Insight 1: The problem requires grouping access times by employee and then checking for overlapping intervals (within a one-hour range).
Insight 2: Sorting the access times for each employee is crucial to efficiently check for the one-hour period constraint.
Insight 3: Converting the time strings to integers simplifies comparison and calculations.

Complexity Analysis

Time Complexity: O(nlogn)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Sorting.

Companies

Asked at: Atlassian.

High-Access Employees | LeetCode 2933

Medium

Array Hash Table String Sorting

Time: O(nlogn)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> findHighAccessEmployees(List<List<String>> access_times) {
        Map<String, List<Integer>> employee_times = new HashMap<>();
        for (List<String> access : access_times) {
            String name = access.get(0);
            String time_str = access.get(1);
            employee_times.computeIfAbsent(name, k -> new ArrayList<>()).add(Integer.parseInt(time_str));
        }

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

            if (times.size() >= 3) {
                for (int i = 0; i <= times.size() - 3; i++) {
                    if (times.get(i + 2) - times.get(i) < 100) {
                        high_access_employees.add(name);
                        break;
                    }
                }
            }
        }

        return high_access_employees;
    }
}

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Problem Info

Description

Problem 2933: High-Access Employees - Detailed explanation and solution approaches.

Topics

Array Hash Table String Sorting

Companies

Atlassian

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2933

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Create a hash map (dictionary) to store access times grouped by employee name.
2
Step 2: Iterate through the input `access_times` array and populate the hash map. Convert the access time strings to integers and store them in a list associated with the employee's name.
3
Step 3: Create an empty list to store the names of high-access employees.
4
Step 4: Iterate through the hash map, processing each employee and their access times.
5
Step 5: Sort the access times for each employee in ascending order.
6
Step 6: Check if the employee has at least three access times. If not, skip to the next employee.
7
Step 7: Iterate through the sorted access times using a sliding window of size 3. For each window, check if the difference between the earliest and latest access time is less than 100 (one hour).
8
Step 8: If the difference is less than 100, it means there are three access times within one hour. Add the employee's name to the list of high-access employees and break out of the inner loop (as the employee is confirmed to be high-access).
9
Step 9: Return the list of high-access employees.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is primarily determined by the sorting step. Each employee's access times are sorted. The hash map stores at most 'n' entries. Iterating through and creating the hashmap is O(n). The nested loop for checking the one-hour window iterates a maximum of 'n' times but this is dominated by the sorting time. Therefore, O(n log n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Approach

Instead of sorting, one could iterate through all possible triplets of access times for each employee and check if they fall within the one-hour window.

Time: O(n^3), where n is the maximum number of access times for an employee. This is because for each employee, all possible triplets would have to be checked.Space: O(n), for storing the hash map of employee access times.

Trade-offs: The brute-force approach is simpler to implement but much less efficient than the sorting-based approach. It should be avoided for larger input sizes.

✨

Key Insights

1Insight 1: The problem requires grouping access times by employee and then checking for overlapping intervals (within a one-hour range).
2Insight 2: Sorting the access times for each employee is crucial to efficiently check for the one-hour period constraint.
3Insight 3: Converting the time strings to integers simplifies comparison and calculations.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The code correctly handles this case because the loops will simply not execute, and an empty list will be returned.
Edge case 2: Employees with fewer than three access times. The code explicitly checks for this condition and skips processing such employees.
Edge case 3: Duplicate access times for the same employee. The code correctly handles duplicate access times as the sorting and the sliding window will identify if there are at least three accesses within the one hour period regardless of duplicates.
Edge case 4: Access times spanning across midnight (e.g., 2350 and 0010). The problem states that these times are not considered part of the same one-hour period, and since the access times are treated as integers and the comparison is `times[i+2] - times[i] < 100`, the solution inherently handles this correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the access times for each employee. Without sorting, it becomes much more difficult to efficiently check for the one-hour period constraint.

✗Common mistake 2: Incorrectly calculating the time difference between access times. The problem states that the times are four digits and should be treated as integers, so a simple subtraction is sufficient. Trying to split the string into hours and minutes and then calculating the difference would add complexity and risk errors.

✗Common mistake 3: Failing to consider that the same high-access employee can have several different one-hour periods that would qualify them as a high-access employee. Only one qualifying one-hour period is sufficient for the inclusion in the returned list.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Security monitoring: Identifying users who are accessing a system frequently within a short period, which could indicate suspicious activity.

🏦Real-world scenario 2: Resource allocation: Determining employees who are heavily utilizing certain resources within a specific timeframe, helping in optimizing resource distribution.

🎮Real-world scenario 3: Anomaly detection: Identifying unusual access patterns that deviate from normal employee behavior, indicating potential security breaches or policy violations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "High-Access Employees"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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