3207. Maximum Points After Enemy Battles LeetCode Solution

In this guide, you will get 3207. Maximum Points After Enemy Battles LeetCode Solution with the best time and space complexity. The solution to Maximum Points After Enemy Battles problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Maximum Points After Enemy Battles solution in C++
Maximum Points After Enemy Battles solution in Java
Maximum Points After Enemy Battles solution in Python
Additional Resources

3207. Maximum Points After Enemy Battles LeetCode Solution image

Problem Statement of Maximum Points After Enemy Battles

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

Complexity Analysis

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

3207. Maximum Points After Enemy Battles LeetCode Solution in C++

class Solution {
 public:
  long long maximumPoints(vector<int>& enemyEnergies, int currentEnergy) {
    const int minEnergy = ranges::min(enemyEnergies);
    return currentEnergy < minEnergy
               ? 0
               : (currentEnergy +
                  accumulate(enemyEnergies.begin(), enemyEnergies.end(), 0LL) -
                  minEnergy) /
                     minEnergy;
  }
};
/* code provided by PROGIEZ */

3207. Maximum Points After Enemy Battles LeetCode Solution in Java

class Solution {
  public long maximumPoints(int[] enemyEnergies, int currentEnergy) {
    final int minEnergy = Arrays.stream(enemyEnergies).min().orElse(0);
    return currentEnergy < minEnergy
        ? 0
        : (currentEnergy + Arrays.stream(enemyEnergies).asLongStream().sum() - minEnergy) /
              minEnergy;
  }
}
// code provided by PROGIEZ

3207. Maximum Points After Enemy Battles LeetCode Solution in Python

class Solution:
  def maximumPoints(self, enemyEnergies: list[int], currentEnergy: int) -> int:
    minEnergy = min(enemyEnergies)
    return (0 if currentEnergy < minEnergy
            else (currentEnergy + sum(enemyEnergies) - minEnergy) // minEnergy)
 # code by PROGIEZ