What is the time complexity of Two Best Non-Overlapping Events?

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

What data structures are used to solve Two Best Non-Overlapping Events?

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

What are the key insights for solving Two Best Non-Overlapping Events?

Insight 1: Sorting events by their start times allows us to process them in a sequential manner and efficiently determine potential non-overlapping event pairs. Insight 2: Using a min-heap (priority queue) helps in tracking the events that have already started but haven't ended yet, ordered by their end times. This allows us to efficiently find the maximum value of events that have already ended when considering a new event.

Which companies ask LeetCode 2054. Two Best Non-Overlapping Events in interviews?

This problem is frequently asked at razorpay.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Two Best Non-Overlapping Events - LeetCode 2054 Solution

Q: How to solve Two Best Non-Overlapping Events?

The solution uses a combination of sorting and a min-heap to efficiently find the maximum sum of values from two non-overlapping events. First, the events are sorted by their start times. Then, we iterate through the sorted events. For each event, we use a min-heap to keep track of the events that have already started but haven't ended. We remove events from the min-heap whose end times are less than the start time of the current event. While removing elements from the min heap, we keep track of the maximum value seen so far in `max_val_prefix`. The maximum sum is then updated as the maximum of the current maximum sum and the sum of the current event's value and the `max_val_prefix`. Finally, the current event is added to the min-heap.

Two Best Non-Overlapping Events - Complete Solution Guide

Two Best Non-Overlapping Events is LeetCode problem 2054, 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 0-indexed 2D integer array of events where events[i] = [startTime i , endTime i , value i ] . The i th event starts at startTime i and ends at endTime i , and if you attend this event, you will receive a value of value i . You can choose at most two non-overlapping events to attend such that the sum of their values is maximized . Return this maximum sum. Note that the start time and end time is inclusive : that is, you cannot attend two events where one of them starts and the oth

Detailed Explanation

The problem asks us to find the maximum sum of values that can be obtained by attending at most two non-overlapping events. Each event has a start time, an end time, and a value. Two events are considered non-overlapping if the end time of the first event is strictly less than the start time of the second event (end_time1 < start_time2). The input is a 2D array where each row represents an event with [startTime, endTime, value]. The output should be the maximum sum of values from at most two non-overlapping events.

Solution Approach

The solution uses a combination of sorting and a min-heap to efficiently find the maximum sum of values from two non-overlapping events. First, the events are sorted by their start times. Then, we iterate through the sorted events. For each event, we use a min-heap to keep track of the events that have already started but haven't ended. We remove events from the min-heap whose end times are less than the start time of the current event. While removing elements from the min heap, we keep track of the maximum value seen so far in `max_val_prefix`. The maximum sum is then updated as the maximum of the current maximum sum and the sum of the current event's value and the `max_val_prefix`. Finally, the current event is added to the min-heap.

Step-by-Step Algorithm

Step 1: Sort the events by their start times in ascending order.
Step 2: Initialize `max_sum` and `max_val_prefix` to 0. Also, initialize an empty min-heap.
Step 3: Iterate through the sorted events.
Step 4: For each event, remove all events from the min-heap that have ended before the current event starts. Update `max_val_prefix` while removing elements.
Step 5: Update `max_sum` to be the maximum of its current value and `max_val_prefix + value` of current event.
Step 6: Add the current event's end time and value to the min-heap.
Step 7: Return `max_sum`.

Key Insights

Insight 1: Sorting events by their start times allows us to process them in a sequential manner and efficiently determine potential non-overlapping event pairs.
Insight 2: Using a min-heap (priority queue) helps in tracking the events that have already started but haven't ended yet, ordered by their end times. This allows us to efficiently find the maximum value of events that have already ended when considering a new event.
Insight 3: Maintaining a `max_val_prefix` variable keeps track of the maximum value among all the events that have already ended. This helps in quickly calculating the potential sum of the current event with a non-overlapping event that has already finished.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Sorting, Heap (Priority Queue).

Companies

Asked at: razorpay.

Two Best Non-Overlapping Events | LeetCode 2054

Medium

Array Binary Search Dynamic Programming 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 maxTwoEvents(int[][] events) {
        Arrays.sort(events, (a, b) -> a[0] - b[0]);
        
        int max_sum = 0;
        int max_val_prefix = 0;
        PriorityQueue<int[]> min_heap = new PriorityQueue<>((a, b) -> a[0] - b[0]);
        
        for (int[] event : events) {
            int start = event[0];
            int end = event[1];
            int value = event[2];
            
            while (!min_heap.isEmpty() && min_heap.peek()[0] < start) {
                int ended_value = min_heap.poll()[1];
                max_val_prefix = Math.max(max_val_prefix, ended_value);
            }
            
            max_sum = Math.max(max_sum, max_val_prefix + value);
            
            min_heap.offer(new int[]{end, value});
        }
        
        return max_sum;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2054: Two Best Non-Overlapping Events - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Sorting Heap (Priority Queue)

Companies

razorpay

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2054

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

7 steps

1
Step 1: Sort the events by their start times in ascending order.
2
Step 2: Initialize `max_sum` and `max_val_prefix` to 0. Also, initialize an empty min-heap.
3
Step 3: Iterate through the sorted events.
4
Step 4: For each event, remove all events from the min-heap that have ended before the current event starts. Update `max_val_prefix` while removing elements.
5
Step 5: Update `max_sum` to be the maximum of its current value and `max_val_prefix + value` of current event.
6
Step 6: Add the current event's end time and value to the min-heap.
7
Step 7: Return `max_sum`.

💡

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 the events takes O(n log n) time. Iterating through the events takes O(n) time. The min-heap operations (insertion and deletion) take O(log n) time each. Since we perform these operations at most n times, the overall time complexity is O(n log n). The space complexity is O(n) because the min-heap can contain at most n elements.

📖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

We can use dynamic programming to solve this problem. Sort the events by end time. Define `dp[i]` as the maximum value achievable by considering events up to index `i`. For each event `i`, we can either include it or not. If we include it, we need to find the latest non-overlapping event before `i` using binary search.

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

Trade-offs: This approach has same complexities, but the heap approach is more intuitive and slightly easier to implement.

✨

Key Insights

1Insight 1: Sorting events by their start times allows us to process them in a sequential manner and efficiently determine potential non-overlapping event pairs.
2Insight 2: Using a min-heap (priority queue) helps in tracking the events that have already started but haven't ended yet, ordered by their end times. This allows us to efficiently find the maximum value of events that have already ended when considering a new event.
3Insight 3: Maintaining a `max_val_prefix` variable keeps track of the maximum value among all the events that have already ended. This helps in quickly calculating the potential sum of the current event with a non-overlapping event that has already finished.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

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

Edge Cases

Edge case 1: The input contains only one event. The algorithm will correctly return the value of that event, as max_val_prefix will be 0.
Edge case 2: All events overlap. The algorithm will still correctly return the maximum value of a single event, as `max_val_prefix` will be updated accordingly.
Edge case 3: Input contains only two non-overlapping events. The algorithm will find these two events and return their sum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the events. Sorting is crucial for efficiently finding non-overlapping events.

✗Common mistake 2: Not using a min-heap. Using a linear search to find the maximum value of a non-overlapping event would increase the time complexity to O(n^2).

✗Common mistake 3: Incorrectly implementing the heap operations, resulting in incorrect results.

✅

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 in a conference room. We want to maximize the total value of the meetings we can schedule, given that some meetings might overlap.

🏦Real-world scenario 2: Optimizing advertisement placements on a website. Each advertisement has a start time, end time, and value (e.g., revenue generated). We want to select the set of non-overlapping advertisements that maximize the total revenue.

💡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: August 14, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Two Best Non-Overlapping Events"?

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

🤖 Need help with this problem?