What is the time complexity of Maximum Number of Events That Can Be Attended?

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

What data structures are used to solve Maximum Number of Events That Can Be Attended?

This problem uses Array, Greedy, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Number of Events That Can Be Attended?

Insight 1: Sorting the events by start day allows us to process events in chronological order, considering only events that are available on a given day. Insight 2: Using a min-heap (priority queue) to store the end days of available events allows us to efficiently choose the event that ends earliest, maximizing the number of days available to attend future events.

Which companies ask LeetCode 1353. Maximum Number of Events That Can Be Attended in interviews?

This problem is frequently asked at Instacart, PayPal, Snowflake, Visa.

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Maximum Number of Events That Can Be Attended - LeetCode 1353 Solution

Q: How to solve Maximum Number of Events That Can Be Attended?

The solution uses a greedy approach combined with a min-heap. We first sort the events by their start day. Then, we iterate through each day, adding the end days of events that start on that day to the min-heap. We remove expired events from the min-heap (events whose end day is less than the current day). Finally, if the min-heap is not empty, we attend the event that ends earliest (the top of the min-heap) and increment the attended event count.

Maximum Number of Events That Can Be Attended - Complete Solution Guide

Maximum Number of Events That Can Be Attended is LeetCode problem 1353, 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 an array of events where events[i] = [startDay i , endDay i ] . Every event i starts at startDay i and ends at endDay i . You can attend an event i at any day d where startDay i <= d <= endDay i . You can only attend one event at any time d . Return the maximum number of events you can attend . Example 1: Input: events = [[1,2],[2,3],[3,4]] Output: 3 Explanation: You can attend all the three events. One way to attend them all is as shown. Attend the first event on day 1. Attend the

Detailed Explanation

The problem asks us to find the maximum number of events we can attend given a list of events, where each event has a start and end day. We can only attend one event per day, and we can only attend an event on a day between its start and end day (inclusive). The goal is to choose a schedule that maximizes the number of events attended.

Solution Approach

The solution uses a greedy approach combined with a min-heap. We first sort the events by their start day. Then, we iterate through each day, adding the end days of events that start on that day to the min-heap. We remove expired events from the min-heap (events whose end day is less than the current day). Finally, if the min-heap is not empty, we attend the event that ends earliest (the top of the min-heap) and increment the attended event count.

Step-by-Step Algorithm

Step 1: Sort the events by their start day in ascending order.
Step 2: Initialize a min-heap (priority queue) to store the end days of available events.
Step 3: Initialize a counter 'attended_count' to 0.
Step 4: Initialize an event index 'event_idx' to 0.
Step 5: Iterate through each day 'd' from 1 up to the maximum possible end day (100000, as per the constraints).
Step 6: While there are events starting on day 'd', add their end days to the min-heap.
Step 7: Remove events from the min-heap that have already ended (end day < d).
Step 8: If the min-heap is not empty, it means there's at least one event available. Attend the event that ends earliest (pop the minimum end day from the min-heap) and increment 'attended_count'.
Step 9: After iterating through all days, return the 'attended_count'.

Key Insights

Insight 1: Sorting the events by start day allows us to process events in chronological order, considering only events that are available on a given day.
Insight 2: Using a min-heap (priority queue) to store the end days of available events allows us to efficiently choose the event that ends earliest, maximizing the number of days available to attend future events.
Insight 3: Removing expired events from the heap on each day ensures we only consider events that are still valid for attendance, preventing us from wasting days on events that have already ended.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Sorting, Heap (Priority Queue).

Companies

Asked at: Instacart, PayPal, Snowflake, Visa.

Maximum Number of Events That Can Be Attended | LeetCode 1353

Medium

Array Greedy Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int maxEvents(int[][] events) {
        // Sort events by their start day
        Arrays.sort(events, (a, b) -> Integer.compare(a[0], b[0]));

        PriorityQueue<Integer> min_heap = new PriorityQueue<>(); // A min-heap to store end days of available events
        int attended_count = 0;
        int event_idx = 0;
        int n = events.length;

        // Iterate through each day from 1 up to the maximum possible day
        for (int d = 1; d <= 100000; d++) {
            // Optimization: if all events are processed and heap is empty, we can stop
            if (event_idx == n && min_heap.isEmpty()) {
                break;
            }

            // Add the end days of all events that start today into the min_heap
            while (event_idx < n && events[event_idx][0] == d) {
                min_heap.offer(events[event_idx][1]);
                event_idx++;
            }

            // Remove events from the min_heap that have already ended
            while (!min_heap.isEmpty() && min_heap.peek() < d) {
                min_heap.poll();
            }

            // If there are any available events, attend the one that ends the soonest
            if (!min_heap.isEmpty()) {
                min_heap.poll();
                attended_count++;
            }
        }

        return attended_count;
    }
}

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Problem Info

Description

Problem 1353: Maximum Number of Events That Can Be Attended - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)

Companies

InstacartPayPalSnowflakeVisa

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1353

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: Sort the events by their start day in ascending order.
2
Step 2: Initialize a min-heap (priority queue) to store the end days of available events.
3
Step 3: Initialize a counter 'attended_count' to 0.
4
Step 4: Initialize an event index 'event_idx' to 0.
5
Step 5: Iterate through each day 'd' from 1 up to the maximum possible end day (100000, as per the constraints).
6
Step 6: While there are events starting on day 'd', add their end days to the min-heap.
7
Step 7: Remove events from the min-heap that have already ended (end day < d).
8
Step 8: If the min-heap is not empty, it means there's at least one event available. Attend the event that ends earliest (pop the minimum end day from the min-heap) and increment 'attended_count'.
9
Step 9: After iterating through all days, return the 'attended_count'.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting takes O(n log n). The outer loop iterates a maximum of 100000 times (based on the constraint 1 <= endDay <= 10^5). Inside the loop, we have heap operations which each take O(log n). The 'while' loops contribute to O(n log n) in total since in the worst case, we could be adding/removing all events to the heap. Space complexity is O(n) as we store all the events' end dates in the heap in the worst case.

📖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

Dynamic Programming

Dynamic programming can be used to solve this problem, but it is generally less efficient than the greedy approach. We can define dp[i] as the maximum number of events that can be attended up to day i. The transitions would involve iterating through all events and checking if attending an event ending on day i yields a better result than not attending any event on that day.

Time: O(n * max_end_day), where n is the number of events and max_end_day is the largest end day.Space: O(max_end_day)

Trade-offs: DP is less efficient because of the O(max_end_day) factor. The greedy approach is more efficient for larger end day ranges as it depends on `n` events instead.

✨

Key Insights

1Insight 1: Sorting the events by start day allows us to process events in chronological order, considering only events that are available on a given day.
2Insight 2: Using a min-heap (priority queue) to store the end days of available events allows us to efficiently choose the event that ends earliest, maximizing the number of days available to attend future events.
3Insight 3: Removing expired events from the heap on each day ensures we only consider events that are still valid for attendance, preventing us from wasting days on events that have already ended.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

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

Edge Cases

Edge case 1: Overlapping events. The greedy approach of picking the event that ends earliest is optimal for handling overlapping events, as it leaves more time for attending other events.
Edge case 2: No events. If the input 'events' is empty, the algorithm will correctly return 0, as 'attended_count' will never be incremented.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the events. Sorting by start day is crucial for the greedy approach to work correctly. Without sorting, we might consider events that haven't started yet, leading to incorrect decisions.

✗Common mistake 2: Not using a min-heap. Using a regular list or array instead of a min-heap to store end days would lead to O(n) time complexity for finding the event that ends earliest, increasing the overall time complexity to O(n^2).

✅

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: Scheduling meetings or appointments. The algorithm can be used to maximize the number of meetings attended, given their start and end times.

🏦Real-world scenario 2: Resource allocation. The algorithm can be applied to allocate resources (e.g., machines, personnel) to tasks with overlapping time windows, maximizing the number of tasks completed.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: September 2, 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 14

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

@TechLeadJul 7

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

@DevExpert92Jun 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Maximum Number of Events That Can Be Attended"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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