#3207

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximum Points After Enemy Battles

Solve the "Maximum Points After Enemy Battles" problem by maximizing points through greedy choices with energy validation.

array greedy

Problem Statement

You are given an array of integers, enemyEnergies, where each element represents the energy required to defeat a particular enemy. You also have an integer, currentEnergy, representing the energy you initially have. Your task is to accumulate as many points as possible by defeating enemies. Each time you defeat an enemy, you gain one point, but you lose energy equal to the enemy’s energy value.

The key is that you can only defeat an enemy if your currentEnergy is greater than or equal to the enemy's energy. Once an enemy is defeated, you gain 1 point and your currentEnergy is reduced by the energy of the defeated enemy. The goal is to find the maximum number of points you can earn by efficiently managing your energy and selecting which enemies to defeat.

Examples

Example 1

Input: enemyEnergies = [3,2,2], currentEnergy = 2

Output: 3

The following operations can be performed to get 3 points, which is the maximum:

Example 2

Input: enemyEnergies = [2] , currentEnergy = 10

Output: 5

Performing the first operation 5 times on enemy 0 results in the maximum number of points.

Constraints

1 <= enemyEnergies.length <= 105
1 <= enemyEnergies[i] <= 109
0 <= currentEnergy <= 109

Solution Approach

Greedy Choice

The solution involves making a greedy choice by defeating enemies with the lowest energy requirements first. By always selecting the enemy that costs the least energy, we maximize the number of points we can achieve with the available energy.

Invariant Validation

To maintain the greedy approach's effectiveness, we must ensure that after each choice, the remaining energy and the potential for further battles are correctly updated. This invariant ensures the algorithm remains valid throughout.

Optimization

We can optimize the solution by sorting the enemyEnergies array and iterating over it, defeating enemies one by one as long as there is enough currentEnergy. This allows us to maximize the points earned efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n log n) due to sorting the enemyEnergies array, where n is the length of the array. The space complexity is O(1) if sorting is done in-place, or O(n) if additional space is required for sorting.

What Interviewers Usually Probe

Ability to identify the greedy approach and its applicability to the problem.
Understanding of invariant validation and why it’s necessary for maintaining the greedy approach’s correctness.
Efficient handling of edge cases where currentEnergy is too small to defeat any enemies.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the enemyEnergies array before starting the battle sequence, which could lead to suboptimal choices.
Not handling the case where the currentEnergy is insufficient to defeat any enemy, which may cause an infinite loop or incorrect results.
Overcomplicating the approach with unnecessary checks or calculations that do not contribute to the greedy choice strategy.

Follow-up variants

Change the array to include enemies with varying point rewards, introducing a new layer of complexity to the greedy choice strategy.
Introduce multiple energy pools that can be used across different rounds of battles, requiring the management of energy distribution.
Modify the problem to include a time-based factor where enemies regenerate energy after each round, adding a dynamic element to the greedy approach.

FAQ

How do I maximize points in "Maximum Points After Enemy Battles"?

Maximize points by following a greedy approach: defeat enemies starting with the least energy requirement, ensuring that you can continue to battle and accumulate points.

What is the primary approach for solving this problem?

The problem is best solved using a greedy strategy, where you always choose the enemy with the smallest energy requirement that you can defeat with your currentEnergy.

What common mistakes should I avoid when solving this problem?

Avoid failing to sort the enemies by energy, not checking for sufficient energy before each battle, and overcomplicating the logic when a simple greedy approach suffices.

Can this problem be solved with dynamic programming?

Although dynamic programming could be applied, this problem is most efficiently solved with a greedy strategy, as it requires the most straightforward approach to maximize points with limited resources.

What are the time and space complexities for the best solution?

The best solution has a time complexity of O(n log n) due to sorting, and a space complexity of O(1) if in-place sorting is used, or O(n) for auxiliary space during sorting.

terminal

Solution

Solution 1: Greedy + Sorting

According to the problem description, we need to score by defeating enemies with the lowest energy value and increase our energy value by defeating enemies with the highest energy value and marking them.

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class Solution:
    def maximumPoints(self, enemyEnergies: List[int], currentEnergy: int) -> int:
        enemyEnergies.sort()
        if currentEnergy < enemyEnergies[0]:
            return 0
        ans = 0
        for i in range(len(enemyEnergies) - 1, -1, -1):
            ans += currentEnergy // enemyEnergies[0]
            currentEnergy %= enemyEnergies[0]
            currentEnergy += enemyEnergies[i]
        return ans

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