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.
\n\nYou can perform either of the following operations zero or multiple times to gain points:
\n\ni, such that currentEnergy >= enemyEnergies[i]. By choosing this option:\n\n\tcurrentEnergy = currentEnergy - enemyEnergies[i].i. By choosing this option:\n\tcurrentEnergy = currentEnergy + enemyEnergies[i].i is marked.Return an integer denoting the maximum points you can get in the end by optimally performing operations.
\n\n\n
Example 1:
\n\nInput: enemyEnergies = [3,2,2], currentEnergy = 2
\n\nOutput: 3
\n\nExplanation:
\n\nThe following operations can be performed to get 3 points, which is the maximum:
\n\npoints increases by 1, and currentEnergy decreases by 2. So, points = 1, and currentEnergy = 0.currentEnergy increases by 3, and enemy 0 is marked. So, points = 1, currentEnergy = 3, and marked enemies = [0].points increases by 1, and currentEnergy decreases by 2. So, points = 2, currentEnergy = 1, and marked enemies = [0].currentEnergy increases by 2, and enemy 2 is marked. So, points = 2, currentEnergy = 3, and marked enemies = [0, 2].points increases by 1, and currentEnergy decreases by 2. So, points = 3, currentEnergy = 1, and marked enemies = [0, 2].Example 2:
\n\nInput: enemyEnergies = [2], currentEnergy = 10
\n\nOutput: 5
\n\nExplanation:
\n\nPerforming the first operation 5 times on enemy 0 results in the maximum number of points.
\n\n
Constraints:
\n\n1 <= enemyEnergies.length <= 1051 <= enemyEnergies[i] <= 1090 <= currentEnergy <= 109Compiled with clang 17 using the latest C++ 23 standard, and libstdc++ provided by GCC 13.
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