How do you solve Finding the Users Active Minutes on LeetCode?

Finding the Users Active Minutes (LeetCode #1817) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient tracking of unique minutes.

What is the brute force solution for Finding the Users Active Minutes?

The brute force approach for Finding the Users Active Minutes is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will iterate through each log entry and count the unique minutes for each user. This is straightforward but inefficient as it checks every log against every other log. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Finding the Users Active Minutes?

Finding the Users Active Minutes uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Finding the Users Active Minutes?

Always clarify the problem requirements before jumping into coding. Think about edge cases, such as users with no actions or all actions in the same minute. Explain your thought process as you code; it shows your understanding.

What are common mistakes when solving Finding the Users Active Minutes?

Not using a set to track unique minutes, leading to incorrect UAM counts. Failing to initialize the answer array correctly.

#1817

Finding the Users Active Minutes

Medium

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a hash map to efficiently track unique minutes for each user, allowing us to compute the user active minutes in linear time.

⚙️

Algorithm

4 steps

1Step 1: Initialize a hash map to store unique minutes for each user.
2Step 2: Iterate through the logs and populate the hash map with unique minutes for each user.
3Step 3: Create an answer array of size k initialized to zero.
4Step 4: Count the number of users for each unique active minute and populate the answer array.

solution.py13 lines

1# Full working Python code
2from collections import defaultdict
3
4def findingUsersActiveMinutes(logs, k):
5    user_minutes = defaultdict(set)
6    for user_id, time in logs:
7        user_minutes[user_id].add(time)
8    
9    answer = [0] * k
10    for minutes in user_minutes.values():
11        uam = len(minutes)
12        answer[uam - 1] += 1
13    return answer

ℹ

Complexity note: The time complexity is O(n) because we process each log entry once. The space complexity is O(n) due to the storage of unique minutes for each user.

1Using a hash map allows for efficient tracking of unique minutes.
2Counting users based on their unique active minutes can be done in a single pass after building the map.

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