Is Minimum Time to Kill All Monsters easy or hard?

Minimum Time to Kill All Monsters is classified as Hard on LeetCode. The difficulty comes from recognizing that the order of actions affects the result and converting a factorial permutation search into an efficient O(n * 2^n) bitmask DP solution.

Minimum Time to Kill All Monsters Python/Java solution

Most implementations use bitmask DP or DFS with memoization. The logic is identical across languages: iterate through all masks, compute the current weapon power using popcount, and update dp[nextMask] with ceil(power[i] / currentPower). Python, Java, C++, Go, and TypeScript implementations follow the same transition pattern.

How to solve Minimum Time to Kill All Monsters in O(n * 2^n)?

Use bitmask dynamic programming. Let dp[mask] represent the minimum time to defeat the monsters in that subset. Count how many monsters are already defeated using popcount(mask) to determine current weapon power, then try defeating each remaining monster and update the next mask with the computed time cost.

What is the best approach for Minimum Time to Kill All Monsters?

Bitmask dynamic programming is the most effective approach. Represent the set of defeated monsters as a bitmask and compute the minimum time for each state. For every state, try killing each remaining monster and update the next state. This reduces the permutation search from O(n!) to O(n * 2^n).

Is Minimum Time to Kill All Monsters asked at Google/Amazon/Meta?

Problems involving bitmask dynamic programming and permutation optimization appear frequently in interviews at companies like Google, Amazon, and Meta. This problem tests understanding of state compression, subset transitions, and DP over subsets.

What data structure is used in Minimum Time to Kill All Monsters?

The core data structure is a bitmask representing subsets of monsters. The algorithm also uses a DP array or memoization map indexed by mask to store the minimum time required for each subset state.

What is the time complexity of Minimum Time to Kill All Monsters?

The optimal solution runs in O(n * 2^n) time and O(2^n) space. There are 2^n possible subsets of monsters, and for each subset the algorithm may try up to n transitions when selecting the next monster to defeat.

Description Hints 3 Solution Video 16:28

2403. Minimum Time to Kill All Monsters

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You are given an integer array power where power[i] is the power of the i^th monster.

You start with 0 mana points, and each day you increase your mana points by gain where gain initially is equal to 1.

Each day, after gaining gain mana, you can defeat a monster if your mana points are greater than or equal to the power of that monster. When you defeat a monster:

your mana points will be reset to 0, and
the value of gain increases by 1.

Return the minimum number of days needed to defeat all the monsters.

Example 1:

Input: power = [3,1,4]
Output: 4
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 2^nd monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points.
- Day 3: Gain 2 mana points to get a total of 4 mana points. Spend all mana points to kill the 3^rd monster.
- Day 4: Gain 3 mana points to get a total of 3 mana points. Spend all mana points to kill the 1^st monster.
It can be proven that 4 is the minimum number of days needed.

Example 2:

Input: power = [1,1,4]
Output: 4
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 1^st monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points. Spend all mana points to kill the 2^nd monster.
- Day 3: Gain 3 mana points to get a total of 3 mana points.
- Day 4: Gain 3 mana points to get a total of 6 mana points. Spend all mana points to kill the 3^rd monster.
It can be proven that 4 is the minimum number of days needed.

Example 3:

Input: power = [1,2,4,9]
Output: 6
Explanation: The optimal way to beat all the monsters is to:
- Day 1: Gain 1 mana point to get a total of 1 mana point. Spend all mana points to kill the 1st monster.
- Day 2: Gain 2 mana points to get a total of 2 mana points. Spend all mana points to kill the 2nd monster.
- Day 3: Gain 3 mana points to get a total of 3 mana points.
- Day 4: Gain 3 mana points to get a total of 6 mana points.
- Day 5: Gain 3 mana points to get a total of 9 mana points. Spend all mana points to kill the 4th monster.
- Day 6: Gain 4 mana points to get a total of 4 mana points. Spend all mana points to kill the 3rd monster.
It can be proven that 6 is the minimum number of days needed.

Constraints:

1 <= power.length <= 17
1 <= power[i] <= 10⁹

Asked at 1 company

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Hints open one at a time

Hint 1

Each monster can only have two states. They are either alive or dead.

Hint 2

We can use bitmasks to represent every possible combination of alive and dead monsters.

Hint 3

Let dp[mask] represent the minimum number of days needed to reach the state mask.

16:28

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