How do you solve High-Access Employees on LeetCode?

High-Access Employees (LeetCode #2933) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting access times allows for efficient counting using a sliding window.

What is the brute force solution for High-Access Employees?

The brute force approach for High-Access Employees is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check every employee's access times against every other access time to see if they fall within the same one-hour period. This is straightforward but inefficient, as it involves nested loops. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve High-Access Employees?

High-Access Employees uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for High-Access Employees?

Always clarify the problem statement and edge cases before jumping into coding. Think about the efficiency of your solution and how to optimize it. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving High-Access Employees?

Not considering the edge cases where access times are at the start or end of the day. Failing to sort the access times before applying the sliding window technique.

#2933

High-Access Employees

Medium

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

This approach leverages sorting and a sliding window technique to efficiently count access times within one-hour periods. By sorting the access times first, we can easily check for counts without nested loops.

⚙️

Algorithm

3 steps

1Step 1: Create a dictionary to store access times for each employee.
2Step 2: For each employee, sort their access times.
3Step 3: Use a sliding window to count how many access times fall within a one-hour period.

solution.py17 lines

1from collections import defaultdict
2
3def high_access_employees(access_times):
4    access_dict = defaultdict(list)
5    for name, time in access_times:
6        access_dict[name].append(time)
7    high_access = []
8    for name, times in access_dict.items():
9        times.sort()
10        left = 0
11        for right in range(len(times)):
12            while int(times[right]) - int(times[left]) >= 60:
13                left += 1
14            if right - left >= 2:
15                high_access.append(name)
16                break
17    return list(set(high_access))

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the access times for each employee. The sliding window technique runs in O(n), making this approach efficient overall.

1Sorting access times allows for efficient counting using a sliding window.
2Using a dictionary to group access times by employee simplifies the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.