LeetCode #3273 — HARD

Minimum Amount of Damage Dealt to Bob

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an integer power and two integer arrays damage and health, both having length n.

Bob has n enemies, where enemy i will deal Bob damage[i] points of damage per second while they are alive (i.e. health[i] > 0).

Every second, after the enemies deal damage to Bob, he chooses one of the enemies that is still alive and deals power points of damage to them.

Determine the minimum total amount of damage points that will be dealt to Bob before all n enemies are dead.

Example 1:

Input: power = 4, damage = [1,2,3,4], health = [4,5,6,8]

Output: 39

Explanation:

  • Attack enemy 3 in the first two seconds, after which enemy 3 will go down, the number of damage points dealt to Bob is 10 + 10 = 20 points.
  • Attack enemy 2 in the next two seconds, after which enemy 2 will go down, the number of damage points dealt to Bob is 6 + 6 = 12 points.
  • Attack enemy 0 in the next second, after which enemy 0 will go down, the number of damage points dealt to Bob is 3 points.
  • Attack enemy 1 in the next two seconds, after which enemy 1 will go down, the number of damage points dealt to Bob is 2 + 2 = 4 points.

Example 2:

Input: power = 1, damage = [1,1,1,1], health = [1,2,3,4]

Output: 20

Explanation:

  • Attack enemy 0 in the first second, after which enemy 0 will go down, the number of damage points dealt to Bob is 4 points.
  • Attack enemy 1 in the next two seconds, after which enemy 1 will go down, the number of damage points dealt to Bob is 3 + 3 = 6 points.
  • Attack enemy 2 in the next three seconds, after which enemy 2 will go down, the number of damage points dealt to Bob is 2 + 2 + 2 = 6 points.
  • Attack enemy 3 in the next four seconds, after which enemy 3 will go down, the number of damage points dealt to Bob is 1 + 1 + 1 + 1 = 4 points.

Example 3:

Input: power = 8, damage = [40], health = [59]

Output: 320

Constraints:

  • 1 <= power <= 104
  • 1 <= n == damage.length == health.length <= 105
  • 1 <= damage[i], health[i] <= 104
Patterns Used
🎯Greedy→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer power and two integer arrays damage and health, both having length n. Bob has n enemies, where enemy i will deal Bob damage[i] points of damage per second while they are alive (i.e. health[i] > 0). Every second, after the enemies deal damage to Bob, he chooses one of the enemies that is still alive and deals power points of damage to them. Determine the minimum total amount of damage points that will be dealt to Bob before all n enemies are dead.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

4
[1,2,3,4]
[4,5,6,8]

Example 2

1
[1,1,1,1]
[1,2,3,4]

Example 3

8
[40]
[59]

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Step 02

Core Insight

What unlocks the optimal approach

  • Can we use sorting here along with a custom comparator?
  • For any two enemies <code>i</code> and <code>j</code> with damages <code>damage[i]</code> and <code>damage[j]</code>, and time to take each of them down <code>t<sub>i</sub></code> and <code>t<sub>j</sub></code>, when is it better to choose enemy <code>i</code> over enemy <code>j</code> first?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #3273: Minimum Amount of Damage Dealt to Bob
class Enemy {
  public int damage;
  public int timeTakenDown;
  public Enemy(int damage, int timeTakenDown) {
    this.damage = damage;
    this.timeTakenDown = timeTakenDown;
  }
}

class Solution {
  public long minDamage(int power, int[] damage, int[] health) {
    long ans = 0;
    long sumDamage = Arrays.stream(damage).asLongStream().sum();
    Enemy[] enemies = new Enemy[damage.length];

    for (int i = 0; i < damage.length; ++i)
      enemies[i] = new Enemy(damage[i], (health[i] + power - 1) / power);

    // It's better to take down the enemy i first if the damage dealt of taking
    // down i first is less than the damage dealt of taking down j first. So,
    //    damage[i] * t[i] + (t[i] + t[j]) * damage[j] <
    //    damage[j] * t[j] + (t[i] + t[j]) * damage[i]
    // => damage[i] * t[i] + damage[j] * t[i] + damage[j] * t[j] <
    //    damage[j] * t[j] + damage[i] * t[j] + damage[i] * t[i]
    // => damage[j] * t[i] < damage[i] * t[j]
    // => damage[j] / t[j] < damage[i] / t[i]
    Arrays.sort(enemies,
                (a, b)
                    -> Double.compare((double) b.damage / b.timeTakenDown,
                                      (double) a.damage / a.timeTakenDown));

    for (final Enemy enemy : enemies) {
      ans += sumDamage * enemy.timeTakenDown;
      sumDamage -= enemy.damage;
    }

    return ans;
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n log n)
Space
O(1)

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.

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