What is the time complexity of Maximum Points After Enemy Battles?

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

What data structures are used to solve Maximum Points After Enemy Battles?

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

What are the key insights for solving Maximum Points After Enemy Battles?

Insight 1: The optimal strategy involves converting the enemy energies into points as efficiently as possible. The crucial part is determining the effective use of the 'marking' operation to maximize subsequent point-earning potential. Insight 2: The minimum energy among the enemies is the key to converting energy into points. By marking all other enemies we can accumulate energy which we can use with the minimum energy enemy.

Which companies ask LeetCode 3207. Maximum Points After Enemy Battles in interviews?

This problem is frequently asked at Rubrik.

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

Maximum Points After Enemy Battles - LeetCode 3207 Solution

Maximum Points After Enemy Battles - Complete Solution Guide

Maximum Points After Enemy Battles is LeetCode problem 3207, 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 integer array enemyEnergies denoting the energy values of various enemies. You are also given an integer currentEnergy denoting the amount of energy you have initially. You start with 0 points, and all the enemies are unmarked initially. You can perform either of the following operations zero or multiple times to gain points: Choose an unmarked enemy, i , such that currentEnergy >= enemyEnergies[i] . By choosing this option: You gain 1 point. Your energy is reduced by the enemy'

Detailed Explanation

The problem presents a scenario where you have an initial energy level (`currentEnergy`) and a list of enemy energies (`enemyEnergies`). The goal is to maximize the number of points you can earn by battling enemies. There are two possible actions for each unmarked enemy: (1) If your current energy is greater than or equal to the enemy's energy, you can fight and gain 1 point while reducing your energy by the enemy's energy. (2) If you have at least 1 point, you can mark an enemy (which has not been marked) and increase your energy by the enemy's energy. The objective is to find the maximum points achievable through an optimal sequence of these operations.

Solution Approach

The solution calculates the minimum energy among all enemies. If the initial energy is less than this minimum energy, it's impossible to defeat any enemy directly, so the result is 0. Otherwise, the solution calculates the total energy available, which is the initial energy plus the sum of all enemy energies minus the minimum energy. Then it divides the total energy by the minimum energy. This division results in the maximum number of points that can be acquired.

Step-by-Step Algorithm

Step 1: Find the minimum energy value among all enemies.
Step 2: Check if the initial energy is less than the minimum enemy energy. If it is, return 0 as no points can be earned.
Step 3: Calculate the sum of all enemy energies.
Step 4: Calculate the total effective energy by adding the initial energy to the sum of enemy energies and subtracting the minimum energy. This represents the total energy available if we mark all other enemies except one with minimal energy.
Step 5: Divide the total effective energy by the minimum energy. This result represents the maximum possible points achievable by optimally using available energy.
Step 6: Return the number of points calculated in the previous step.

Key Insights

Insight 1: The optimal strategy involves converting the enemy energies into points as efficiently as possible. The crucial part is determining the effective use of the 'marking' operation to maximize subsequent point-earning potential.
Insight 2: The minimum energy among the enemies is the key to converting energy into points. By marking all other enemies we can accumulate energy which we can use with the minimum energy enemy.
Insight 3: If initial energy is less than the minimum enemy energy, no points can be earned.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy.

Companies

Asked at: Rubrik.

Maximum Points After Enemy Battles | LeetCode 3207

Medium

Array Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public long maximumPoints(int[] enemyEnergies, int currentEnergy) {
        if (enemyEnergies.length == 0) {
            return currentEnergy > 0 ? currentEnergy : 0;
        }

        int min_energy = enemyEnergies[0];
        for (int energy : enemyEnergies) {
            min_energy = Math.min(min_energy, energy);
        }

        if (currentEnergy < min_energy) {
            return 0;
        }

        long total_energy = currentEnergy;
        for (int energy : enemyEnergies) {
            total_energy += energy;
        }
        total_energy -= min_energy;

        long points = total_energy / min_energy;

        return points;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3207: Maximum Points After Enemy Battles - Detailed explanation and solution approaches.

Topics

Array Greedy

Companies

Rubrik

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3207

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: Find the minimum energy value among all enemies.
2
Step 2: Check if the initial energy is less than the minimum enemy energy. If it is, return 0 as no points can be earned.
3
Step 3: Calculate the sum of all enemy energies.
4
Step 4: Calculate the total effective energy by adding the initial energy to the sum of enemy energies and subtracting the minimum energy. This represents the total energy available if we mark all other enemies except one with minimal energy.
5
Step 5: Divide the total effective energy by the minimum energy. This result represents the maximum possible points achievable by optimally using available energy.
6
Step 6: Return the number of points calculated in the previous step.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the loop that iterates through the `enemyEnergies` array to find the minimum energy and calculate the sum. Other operations, such as addition, subtraction, and division, take constant time. The space complexity is constant because the algorithm uses only a few variables to store values like `min_energy`, `total_energy`, and `points`.

📖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

A dynamic programming approach could be considered, where dp[i][j] stores the maximum points achievable considering the first i enemies with j energy. However, the large input constraints would result in impractical memory usage.

Time: O(n*E), where n is the number of enemies and E is the total energySpace: O(n*E)

Trade-offs: Dynamic Programming is unlikely to be helpful due to large memory constraints, making it less practical than greedy for this problem.

✨

Key Insights

1Insight 1: The optimal strategy involves converting the enemy energies into points as efficiently as possible. The crucial part is determining the effective use of the 'marking' operation to maximize subsequent point-earning potential.
2Insight 2: The minimum energy among the enemies is the key to converting energy into points. By marking all other enemies we can accumulate energy which we can use with the minimum energy enemy.
3Insight 3: If initial energy is less than the minimum enemy energy, no points can be earned.

📚

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.

Edge Cases

Edge case 1: Empty enemyEnergies array: The C++ and Java solutions correctly handle an empty input array. The Python solution doesn't need explicit handling as the min() function will raise an error, which is acceptable.
Edge case 2: currentEnergy is 0: The algorithm correctly handles this as the total effective energy is still calculated appropriately. If `currentEnergy` is 0, the maximum points will be determined solely by the enemies' energies.
Edge case 3: currentEnergy is less than min_energy: The algorithm handles this by returning 0 because no initial move is possible.
Edge case 4: Large enemy energies: The Java and C++ solutions use `long` to prevent integer overflow when calculating the total energy, ensuring correctness for large input values. The C solution uses long long.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow: Failing to use long data types when summing energies can cause integer overflow, leading to incorrect results, especially with the given constraints.

✗Common mistake 2: Not finding the minimum energy: Overlooking the importance of the minimum energy as a divisor for the total energy can lead to incorrect strategies and suboptimal points.

✗Common mistake 3: Incorrect initial condition for corner case: If the initial energy is less than min energy, students sometimes forget to handle that no moves are possible.

✅

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: Imagine allocating resources (energy) to projects (enemies) with varying costs. Finding the most efficient project (lowest cost) and using resources to complete multiples of it relates to the problem's core concept.

🏦Real-world scenario 2: Investment strategies: Maximizing returns from investments can be likened to maximizing points. Identifying the investment with the lowest entry barrier (minimum enemy energy) and strategically reinvesting profits relates to the concept of energy accumulation.

💡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: October 26, 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.

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 Points After Enemy Battles"?

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?