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3207. Maximum Points After Enemy Battles
Description
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 thatcurrentEnergy >= 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.
 Your energy increases by the enemy's energy, i.e.
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, andcurrentEnergy
decreases by 2. So,points = 1
, andcurrentEnergy = 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, andcurrentEnergy
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, andcurrentEnergy
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 <= 10^{5}
1 <= enemyEnergies[i] <= 10^{9}
0 <= currentEnergy <= 10^{9}
Solutions
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.
Therefore, we can sort the enemies by their energy values, then start from the enemy with the highest energy value, always choose the enemy with the lowest energy value to score and consume energy. Next, we add the energy value of the enemy with the highest energy to our current energy and mark that enemy. Repeat the above steps until all enemies are marked.
The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$, where $n$ is the number of enemies.

class Solution { public long maximumPoints(int[] enemyEnergies, int currentEnergy) { Arrays.sort(enemyEnergies); if (currentEnergy < enemyEnergies[0]) { return 0; } long ans = 0; for (int i = enemyEnergies.length  1; i >= 0; i) { ans += currentEnergy / enemyEnergies[0]; currentEnergy %= enemyEnergies[0]; currentEnergy += enemyEnergies[i]; } return ans; } };

class Solution { public: long long maximumPoints(vector<int>& enemyEnergies, int currentEnergy) { sort(enemyEnergies.begin(), enemyEnergies.end()); if (currentEnergy < enemyEnergies[0]) { return 0; } long long ans = 0; for (int i = enemyEnergies.size()  1; i >= 0; i) { ans += currentEnergy / enemyEnergies[0]; currentEnergy %= enemyEnergies[0]; currentEnergy += enemyEnergies[i]; } return ans; } };

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

func maximumPoints(enemyEnergies []int, currentEnergy int) (ans int64) { sort.Ints(enemyEnergies) if currentEnergy < enemyEnergies[0] { return 0 } for i := len(enemyEnergies)  1; i >= 0; i { ans += int64(currentEnergy / enemyEnergies[0]) currentEnergy %= enemyEnergies[0] currentEnergy += enemyEnergies[i] } return }

function maximumPoints(enemyEnergies: number[], currentEnergy: number): number { enemyEnergies.sort((a, b) => a  b); if (currentEnergy < enemyEnergies[0]) { return 0; } let ans = 0; for (let i = enemyEnergies.length  1; ~i; i) { ans += Math.floor(currentEnergy / enemyEnergies[0]); currentEnergy %= enemyEnergies[0]; currentEnergy += enemyEnergies[i]; } return ans; }