Is Maximum Points After Enemy Battles easy or hard?

Maximum Points After Enemy Battles is generally considered a Medium difficulty problem. The implementation is straightforward once the greedy insight is recognized, but identifying the correct strategy and managing the two-pointer decisions can take some practice.

Maximum Points After Enemy Battles Python/Java solution

Python and Java implementations typically sort the enemy array and maintain two pointers. The algorithm increases points by defeating weaker enemies and conditionally sacrifices a point to regain energy from stronger enemies, ensuring the maximum achievable score.

How to solve Maximum Points After Enemy Battles in O(n log n)?

First sort the enemy energy array. Use two pointers: the left pointer targets the weakest enemy to gain points, while the right pointer targets the strongest enemy to regain energy when needed. Continue defeating weak enemies whenever energy allows and trade points only when it unlocks additional wins.

What is the best approach for Maximum Points After Enemy Battles?

The most efficient approach is a greedy strategy with sorting. Sort enemies by required energy, defeat the weakest enemies first to accumulate points, and if energy runs out, trade a point to regain energy using the strongest remaining enemy. This approach runs in O(n log n) time due to sorting and uses O(1) extra space.

Is Maximum Points After Enemy Battles asked at Google/Amazon/Meta?

Problems based on greedy resource allocation and two-pointer strategies appear frequently in interviews at companies like Amazon, Google, and Meta. This problem closely resembles classic greedy interview questions such as Bag of Tokens, making it relevant preparation.

What data structure is used in Maximum Points After Enemy Battles?

The core data structure is an array combined with sorting and a two-pointer technique. The algorithm repeatedly accesses the smallest and largest remaining values to decide whether to gain points or recover energy.

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

The optimal greedy solution runs in O(n log n) time because the enemy list must be sorted. After sorting, a two-pointer scan processes each enemy at most once. Space complexity is O(1) if sorting is done in place.

Description Hints 4 Solution Video 12:15

3207. Maximum Points After Enemy Battles

Medium33.2% acceptance1 company

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's energy, i.e. currentEnergy = currentEnergy - enemyEnergies[i].
If you have at least 1 point, you can choose an unmarked enemy, i. By choosing this option:
- Your energy increases by the enemy's energy, i.e. currentEnergy = currentEnergy + enemyEnergies[i].
- The enemy i is marked.

Return an integer denoting the maximum points you can get in the end by optimally performing operations.

Example 1:

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

Output: 3

Explanation:

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

First operation on enemy 1: points increases by 1, and currentEnergy decreases by 2. So, points = 1, and currentEnergy = 0.
Second operation on enemy 0: currentEnergy increases by 3, and enemy 0 is marked. So, points = 1, currentEnergy = 3, and marked enemies = [0].
First operation on enemy 2: points increases by 1, and currentEnergy decreases by 2. So, points = 2, currentEnergy = 1, and marked enemies = [0].
Second operation on enemy 2: currentEnergy increases by 2, and enemy 2 is marked. So, points = 2, currentEnergy = 3, and marked enemies = [0, 2].
First operation on enemy 1: points increases by 1, and currentEnergy decreases by 2. So, points = 3, currentEnergy = 1, and marked enemies = [0, 2].

Example 2:

Input: enemyEnergies = [2], currentEnergy = 10

Output: 5

Explanation:

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

Asked at 1 company

Rubrik

Topics

Array Greedy

Hints open one at a time

Hint 1

The problem can be solved greedily.

Hint 2

Mark all the others except the smallest one first.

Hint 3

Use the smallest one to increase the energy.

Hint 4

Note that the initial energy should be no less than the smallest enemy.

12:15

Still stuck?Read 3 approaches, or watch a walkthrough.

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1class Solution:
2    def maximumPoints(self, enemyEnergies: List[int], currentEnergy: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[3,2,2]
2

Input

[2]
10

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