Is Minimum Time to Break Locks II easy or hard?

Minimum Time to Break Locks II is classified as Hard. The difficulty comes from recognizing that the lock-breaking order affects future costs and that brute force permutations are too slow. Converting the problem into subset DP with memoized DFS is the key insight.

Minimum Time to Break Locks II Python/Java solution

Implement a recursive DFS function that accepts the current bitmask of broken locks. Use a memo array indexed by mask to cache results. For each unbroken lock, compute the time using ceil(strength[i] / power), update the mask, and recurse. The same logic works in Python, Java, C++, and Go with bit operations and memoization.

How to solve Minimum Time to Break Locks II in O(n · 2^n)?

Represent the set of broken locks as a bitmask. For each mask, compute the current power using the number of bits set. Try breaking each remaining lock, calculate the time cost using ceil(strength[i] / power), and recursively solve the next state. Cache results for each mask so every subset is evaluated only once.

What is the best approach for Minimum Time to Break Locks II?

The most efficient approach is DFS with memoization using a bitmask to represent which locks are already broken. Each state computes the minimum time to finish the remaining locks while deriving the current sword power from the number of broken locks. This reduces repeated work and runs in O(n · 2^n) time with O(2^n) space.

Is Minimum Time to Break Locks II asked at Google/Amazon/Meta?

Problems involving subset DP, permutation optimization, and state graph traversal frequently appear in interviews at companies like Google, Amazon, and Meta. This problem tests similar skills: modeling states, pruning repeated work, and optimizing exponential search with memoization.

What data structure is used in Minimum Time to Break Locks II?

The core structure is a bitmask representing which locks are already broken. A memoization map or array stores results for each mask. DFS or graph traversal is used to explore transitions between states while computing the minimum cost.

What is the time complexity of Minimum Time to Break Locks II?

A naive permutation search takes O(n!) time because every order of locks must be evaluated. With bitmask dynamic programming and memoization, the complexity improves to O(n · 2^n) since each subset of locks is processed once and transitions try up to n remaining locks.

Description Hints 1 Solution

3385. Minimum Time to Break Locks II

Hard45.7% acceptance1 companyPremium elsewhere, free here

Bob is stuck in a dungeon and must break n locks, each requiring some amount of energy to break. The required energy for each lock is stored in an array called strength where strength[i] indicates the energy needed to break the i^th lock.

To break a lock, Bob uses a sword with the following characteristics:

The initial energy of the sword is 0.
The initial factor X by which the energy of the sword increases is 1.
Every minute, the energy of the sword increases by the current factor X.
To break the i^th lock, the energy of the sword must reach at least strength[i].
After breaking a lock, the energy of the sword resets to 0, and the factor X increases by 1.

Your task is to determine the minimum time in minutes required for Bob to break all n locks and escape the dungeon.

Return the minimum time required for Bob to break all n locks.

Example 1:

Input: strength = [3,4,1]

Output: 4

Explanation:

Time	Energy	X	Action	Updated X
0	0	1	Nothing	1
1	1	1	Break 3^rd Lock	2
2	2	2	Nothing	2
3	4	2	Break 2^nd Lock	3
4	3	3	Break 1^st Lock	3

The locks cannot be broken in less than 4 minutes; thus, the answer is 4.

Example 2:

Input: strength = [2,5,4]

Output: 6

Explanation:

Time	Energy	X	Action	Updated X
0	0	1	Nothing	1
1	1	1	Nothing	1
2	2	1	Break 1^st Lock	2
3	2	2	Nothing	2
4	4	2	Break 3^rd Lock	3
5	3	3	Nothing	3
6	6	3	Break 2^nd Lock	4

The locks cannot be broken in less than 6 minutes; thus, the answer is 6.

Constraints:

n == strength.length
1 <= n <= 80
1 <= strength[i] <= 10⁶
n == strength.length

Asked at 1 company

Ivp

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Hint 1

Use the Hungarian algorithm for minimum cost matching.

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1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[3,4,1]

Input

[2,5,4]

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