What is the time complexity of Finding the Users Active Minutes?

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

What data structures are used to solve Finding the Users Active Minutes?

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

What are the key insights for solving Finding the Users Active Minutes?

Insight 1: We need to track the unique minutes each user has performed an action. Insight 2: Hash tables (dictionaries, maps) are ideal for storing and retrieving user activity data efficiently, allowing us to quickly check if a user has performed an action at a specific minute.

Which companies ask LeetCode 1817. Finding the Users Active Minutes in interviews?

This problem is frequently asked at X.

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Finding the Users Active Minutes - LeetCode 1817 Solution

Finding the Users Active Minutes - Complete Solution Guide

Finding the Users Active Minutes is LeetCode problem 1817, 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 the logs for users' actions on LeetCode, and an integer k . The logs are represented by a 2D integer array logs where each logs[i] = [ID i , time i ] indicates that the user with ID i performed an action at the minute time i . Multiple users can perform actions simultaneously, and a single user can perform multiple actions in the same minute. The user active minutes (UAM) for a given user is defined as the number of unique minutes in which the user performed an action on LeetCode.

Detailed Explanation

The problem asks us to process user activity logs to determine how many users have a specific User Active Minutes (UAM). UAM for a user is the number of unique minutes they performed an action. Given a 2D array 'logs' where each entry is [userId, time] and an integer 'k', we need to return an array 'answer' of size 'k'. `answer[j]` should represent the number of users whose UAM is equal to `j+1`. In other words, we count how many users have exactly 1 active minute, 2 active minutes, up to k active minutes.

Solution Approach

The provided solutions use a hash table (dictionary/map) to store each user's unique active minutes. The algorithm iterates through the logs, adding each minute to the corresponding user's set of minutes. After processing all logs, it counts how many users have a UAM of 1, 2, ..., k, and stores the counts in the 'answer' array.

Step-by-Step Algorithm

Step 1: Initialize a hash table (e.g., dictionary in Python, HashMap in Java, unordered_map in C++) to store user IDs as keys and a set of their unique active minutes as values.
Step 2: Iterate through the 'logs' array. For each log entry [userId, time], add the 'time' to the set associated with the 'userId' in the hash table. Use set's built-in functionality to ensure uniqueness of minutes.
Step 3: Initialize an 'answer' array of size 'k' with all elements set to 0.
Step 4: Iterate through the values (sets of minutes) in the hash table. For each set, calculate its size, which represents the UAM for the corresponding user.
Step 5: If the calculated UAM is less than or equal to 'k', increment the element at index 'UAM - 1' in the 'answer' array. This is because answer[j] corresponds to users with UAM j+1.
Step 6: Return the 'answer' array.

Key Insights

Insight 1: We need to track the unique minutes each user has performed an action.
Insight 2: Hash tables (dictionaries, maps) are ideal for storing and retrieving user activity data efficiently, allowing us to quickly check if a user has performed an action at a specific minute.
Insight 3: The size of the answer array is determined by 'k', which provides the upper bound for UAM. The index 'j' in answer corresponds to UAM j+1.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table.

Companies

Asked at: X.

Finding the Users Active Minutes | LeetCode 1817

Medium

Array Hash Table

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

class Solution {
    public int[] findingUsersActiveMinutes(int[][] logs, int k) {
        Map<Integer, Set<Integer>> userMinutes = new HashMap<>();

        for (int[] log : logs) {
            int userId = log[0];
            int time = log[1];
            userMinutes.computeIfAbsent(userId, key -> new HashSet<>()).add(time);
        }

        int[] answer = new int[k];

        for (Set<Integer> minutesSet : userMinutes.values()) {
            int uam = minutesSet.size();
            if (uam <= k) {
                answer[uam - 1] += 1;
            }
        }

        return answer;
    }
}

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Problem Info

Description

Problem 1817: Finding the Users Active Minutes - Detailed explanation and solution approaches.

Topics

Array Hash Table

Companies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1817

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

6 steps

1
Step 1: Initialize a hash table (e.g., dictionary in Python, HashMap in Java, unordered_map in C++) to store user IDs as keys and a set of their unique active minutes as values.
2
Step 2: Iterate through the 'logs' array. For each log entry [userId, time], add the 'time' to the set associated with the 'userId' in the hash table. Use set's built-in functionality to ensure uniqueness of minutes.
3
Step 3: Initialize an 'answer' array of size 'k' with all elements set to 0.
4
Step 4: Iterate through the values (sets of minutes) in the hash table. For each set, calculate its size, which represents the UAM for the corresponding user.
5
Step 5: If the calculated UAM is less than or equal to 'k', increment the element at index 'UAM - 1' in the 'answer' array. This is because answer[j] corresponds to users with UAM j+1.
6
Step 6: Return the 'answer' array.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the loop that processes the logs. The operations within the loop (hash table lookups and set insertions) take O(1) on average. The final loop iterates through a data structure whose overall size is also bounded by the number of logs. The space complexity is O(n) because the hash table stores, in the worst case, distinct userId-time pairs which is bounded by 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

Sorting and Counting

Sort the logs based on user ID, then iterate through the sorted logs. For each user, keep track of the unique minutes using a temporary set. After processing each user, calculate their UAM and update the answer array.

Time: O(n log n) due to sorting.Space: O(n) in the worst case, for storing the logs and temporary sets.

Trade-offs: Sorting is slower than using a hash table. This approach may be beneficial if the log data is already partially sorted or if the available memory is limited, making the hash table approach infeasible.

✨

Key Insights

1Insight 1: We need to track the unique minutes each user has performed an action.
2Insight 2: Hash tables (dictionaries, maps) are ideal for storing and retrieving user activity data efficiently, allowing us to quickly check if a user has performed an action at a specific minute.
3Insight 3: The size of the answer array is determined by 'k', which provides the upper bound for UAM. The index 'j' in answer corresponds to UAM j+1.

📚

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!

Edge Cases

Edge case 1: Empty 'logs' array. The code handles this correctly by returning an array of all zeros because the loops won't execute.
Edge case 2: Large number of logs for a single user. The solution handles this case by storing the unique minutes for each user using a set. The size of the set for each user is bounded by the number of logs, which contributes to the overall O(n) space complexity.
Edge case 3: k is smaller than the maximum UAM. The code still functions correctly; the UAMs which are greater than k are not added to the result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the uniqueness of minutes for each user. If minutes are not tracked using a set, the UAM will be incorrectly calculated.

✗Common mistake 2: Incorrectly indexing the 'answer' array. The index should be 'UAM - 1' since the 'answer' array is 1-indexed based on UAM, while arrays are generally 0-indexed.

✗Common mistake 3: Not handling the case where the log is empty. The provided solutions correctly handle an empty input log returning the correct zero-initialized answer array

✅

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: Analyzing website user activity to understand how engaged users are. UAM could represent the number of unique days a user visited a website or used a specific feature.

🏦Real-world scenario 2: Monitoring server logs to identify the number of unique minutes a server experienced high traffic or specific error events. This can help in capacity planning and troubleshooting.

💡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: June 18, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

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 "Finding the Users Active Minutes"?

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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