LeetCode #3207 — MEDIUM

Maximum Points After Enemy Battles

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[3,2,2]
2

Example 2

[2]
10

Step 02

Core Insight

What unlocks the optimal approach

The problem can be solved greedily.
Mark all the others except the smallest one first.
Use the smallest one to increase the energy.
Note that the initial energy should be no less than the smallest enemy.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3207: Maximum Points After Enemy Battles
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;
    }
};

// Accepted solution for LeetCode #3207: Maximum Points After Enemy Battles
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
}

# Accepted solution for LeetCode #3207: Maximum Points After Enemy Battles
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

// Accepted solution for LeetCode #3207: Maximum Points After Enemy Battles
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3207: Maximum Points After Enemy Battles
// 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;
//     }
// };

// Accepted solution for LeetCode #3207: Maximum Points After Enemy Battles
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.