What is the time complexity of Eliminate Maximum Number of Monsters?

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

What data structures are used to solve Eliminate Maximum Number of Monsters?

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

What are the key insights for solving Eliminate Maximum Number of Monsters?

Insight 1: The key is to calculate the arrival time for each monster to the city. The arrival time is calculated as distance / speed. Insight 2: Sorting the arrival times is crucial. This allows us to prioritize eliminating monsters that will reach the city sooner.

Which companies ask LeetCode 1921. Eliminate Maximum Number of Monsters in interviews?

This problem is frequently asked at Agoda.

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Eliminate Maximum Number of Monsters - LeetCode 1921 Solution

Eliminate Maximum Number of Monsters - Complete Solution Guide

Eliminate Maximum Number of Monsters is LeetCode problem 1921, 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 playing a video game where you are defending your city from a group of n monsters. You are given a 0-indexed integer array dist of size n , where dist[i] is the initial distance in kilometers of the i th monster from the city. The monsters walk toward the city at a constant speed. The speed of each monster is given to you in an integer array speed of size n , where speed[i] is the speed of the i th monster in kilometers per minute. You have a weapon that, once fully charged, can eliminat

Detailed Explanation

The problem describes a scenario where you are defending a city from monsters. Each monster has an initial distance from the city and a speed. You have a weapon that can eliminate one monster per minute, starting from the very beginning. The goal is to find the maximum number of monsters you can eliminate before any monster reaches the city. If a monster reaches the city at the exact moment you're charging your weapon, you lose. The inputs are two arrays: `dist` representing the initial distance of each monster, and `speed` representing the speed of each monster. The output is the maximum number of monsters that can be eliminated.

Solution Approach

The solution calculates the arrival time for each monster by dividing its distance by its speed. These arrival times are stored in an array and then sorted in ascending order. The code then iterates through the sorted arrival times. In each iteration, it checks if the arrival time of the current monster is less than or equal to the current time (which is also the number of monsters eliminated so far). If it is, it means the monster has reached the city (or is about to reach) and the game is lost. Therefore, we return the number of monsters eliminated up to that point. If all monsters can be eliminated before reaching the city, the function returns the total number of monsters.

Step-by-Step Algorithm

Step 1: Calculate the arrival time for each monster by dividing its distance by its speed. Store these times in an array.
Step 2: Sort the array of arrival times in ascending order.
Step 3: Iterate through the sorted arrival times, with the index `i` representing the current minute (and also the number of monsters eliminated so far).
Step 4: In each iteration, check if the arrival time of the current monster is less than or equal to `i`. If it is, return `i` (the number of monsters eliminated so far).
Step 5: If the loop completes without returning, it means all monsters can be eliminated. Return the total number of monsters `n`.

Key Insights

Insight 1: The key is to calculate the arrival time for each monster to the city. The arrival time is calculated as distance / speed.
Insight 2: Sorting the arrival times is crucial. This allows us to prioritize eliminating monsters that will reach the city sooner.
Insight 3: The time spent eliminating monsters increases by 1 minute with each eliminated monster. We need to make sure that at each time step `i`, the arrival time of the `i`-th monster is greater than `i` to avoid losing.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Agoda.

Eliminate Maximum Number of Monsters | LeetCode 1921

Medium

Array Greedy Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int eliminateMaximum(int[] dist, int[] speed) {
        int n = dist.length;
        double[] arrival_times = new double[n];
        
        for (int i = 0; i < n; i++) {
            arrival_times[i] = (double) dist[i] / speed[i];
        }
        
        Arrays.sort(arrival_times);
        
        for (int i = 0; i < n; i++) {
            if (arrival_times[i] <= i) {
                return i;
            }
        }
        
        return n;
    }
}

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Problem Info

Description

Problem 1921: Eliminate Maximum Number of Monsters - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

Agoda

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1921

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

5 steps

1
Step 1: Calculate the arrival time for each monster by dividing its distance by its speed. Store these times in an array.
2
Step 2: Sort the array of arrival times in ascending order.
3
Step 3: Iterate through the sorted arrival times, with the index `i` representing the current minute (and also the number of monsters eliminated so far).
4
Step 4: In each iteration, check if the arrival time of the current monster is less than or equal to `i`. If it is, return `i` (the number of monsters eliminated so far).
5
Step 5: If the loop completes without returning, it means all monsters can be eliminated. Return the total number of monsters `n`.

💡

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?

The time complexity is dominated by the sorting step, which takes O(n log n) time. The rest of the algorithm takes O(n) time. Therefore, the overall time complexity is O(n log n). The space complexity is O(n) because we create a new array to store the arrival times.

📖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

Heap (Priority Queue)

Instead of sorting, use a min-heap to store arrival times. In each iteration, extract the monster with the smallest arrival time. If the arrival time is less than or equal to the current time, return the number of monsters eliminated so far.

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

Trade-offs: Heap-based approach has the same time complexity of O(n log n). It might offer a slight performance difference in specific scenarios but generally isn't a significant improvement over sorting for this problem. It adds more code complexity for same time complexity.

✨

Key Insights

1Insight 1: The key is to calculate the arrival time for each monster to the city. The arrival time is calculated as distance / speed.
2Insight 2: Sorting the arrival times is crucial. This allows us to prioritize eliminating monsters that will reach the city sooner.
3Insight 3: The time spent eliminating monsters increases by 1 minute with each eliminated monster. We need to make sure that at each time step `i`, the arrival time of the `i`-th monster is greater than `i` to avoid losing.

📚

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: When multiple monsters have the same arrival time. Sorting ensures that the monsters are processed from earliest to latest arrival time. The solution correctly handles multiple monsters arriving at the same time.
Edge case 2: When all monsters have a large distance and small speed, resulting in very large arrival times. The solution will still correctly calculate and sort the arrival times and return the total number of monsters.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not calculating the arrival time correctly (distance / speed). It is crucial to use floating-point division to avoid integer truncation.

✗Common mistake 2: Not sorting the arrival times. Sorting is essential for prioritizing monsters that will reach the city sooner.

✗Common mistake 3: Forgetting to consider the case where all monsters can be eliminated. The code must return `n` if the loop completes successfully.

✅

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: Resource allocation and task scheduling. Imagine allocating resources to tasks with different deadlines and processing times. This problem's approach could help determine how many tasks can be completed before a critical deadline is missed.

🏦Real-world scenario 2: Network packet processing. Packets arrive at a router with different priorities and processing times. The router needs to process packets in a way that maximizes the number of packets processed before a buffer overflow occurs. This algorithm can model prioritizing packet processing based on deadlines or priorities.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 30, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 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 "Eliminate Maximum Number of Monsters"?

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.

🤖 Need help with this problem?