LeetCode 3376 - Minimum Time to Break Locks I

The problem describes a sequence of locks that must all be broken. Each lock requires a certain amount of energy, given by strength[i]. Bob's sword starts with: - Energy = 0 - Growth factor x = 1 Every minute, the sword gains x energy.

leetcodemediumarraydynamic-programmingbacktrackingbit-manipulationbreadth-first-searchbitmask

LeetCode Problem 3376

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Backtracking, Bit Manipulation, Breadth-First Search, Bitmask

Solution

LeetCode 3376 - Minimum Time to Break Locks I

Problem Understanding

The problem describes a sequence of locks that must all be broken. Each lock requires a certain amount of energy, given by strength[i].

Bob's sword starts with:

  • Energy = 0
  • Growth factor x = 1

Every minute, the sword gains x energy.

To break a lock whose required strength is s, the current energy must be at least s. Once the lock is broken:

  • The sword's energy immediately resets to 0.
  • The growth factor increases by k.

The important observation is that after breaking some number of locks, the growth factor is completely determined. If we have already broken m locks, then:

$$x = 1 + m \cdot k$$

When the current growth factor is x, breaking a lock with strength s requires waiting:

$$\left\lceil \frac{s}{x} \right\rceil$$

minutes, because energy grows linearly from 0 at rate x per minute.

Therefore, the real decision is the order in which we break the locks. Different orders produce different growth factors when each lock is attempted, which changes the required waiting times.

The input consists of:

  • strength, an array of lock strengths.
  • k, the amount added to the growth factor after each lock.

The output is the minimum total number of minutes required to break every lock.

Constraints Analysis

The key constraint is:

  • n ≤ 8

This is very small.

A naive search over all lock orders would examine:

$$n!$$

permutations.

For n = 8:

$$8! = 40320$$

which is not huge, but it suggests that exponential or factorial approaches may be acceptable.

The small value of n strongly hints at a bitmask dynamic programming solution.

Important Edge Cases

When n = 1, there is only one lock and the answer is simply the number of minutes required to reach its strength with x = 1.

When all strengths are identical, different orders may still matter because later locks benefit from larger growth factors.

When one lock is extremely large and the others are very small, it may be advantageous to delay the large lock until the growth factor has increased substantially.

Since strengths can reach 10^6, any approach that tries to simulate energy minute by minute would be far too slow. We must compute waiting times directly using ceiling division.

Approaches

Brute Force

The most direct approach is to try every possible order of breaking locks.

For a given permutation:

  1. Start with x = 1.
  2. Process locks in that order.
  3. For each lock, add the required waiting time:

$$\left\lceil \frac{strength[i]}{x} \right\rceil$$ 4. Increase x by k. 5. Keep the minimum total time among all permutations.

This is correct because every valid strategy corresponds to some permutation of the locks.

However, it requires checking all n! orders. While 8! = 40320 is manageable, we can do better with dynamic programming.

Key Insight

The only information that matters at any point is which locks have already been broken.

Suppose a set of locks has already been broken.

Then:

  • The number of broken locks is known.
  • Therefore the current growth factor is known.
  • The future cost depends only on the remaining locks.

This means we can define a DP state using a bitmask representing the set of broken locks.

For each state, we try breaking one remaining lock next.

Since there are only:

$$2^n$$

possible subsets and n ≤ 8, a bitmask DP is very efficient.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(n! · n) O(n) Enumerates every lock order
Optimal Bitmask DP O(n · 2^n) O(2^n) Computes each subset state once

Algorithm Walkthrough

Step 1

Represent the set of already broken locks using a bitmask.

If bit i is set, lock i has already been broken.

Step 2

Define:

$$dp(mask)$$

as the minimum additional time needed to break all remaining locks starting from state mask.

Step 3

Count how many locks have already been broken:

$$broken = \text{popcount}(mask)$$

Since every broken lock increases the factor by k:

$$x = 1 + broken \cdot k$$

Step 4

If all locks have been broken:

$$mask = (1<<n)-1$$

then no more time is required, so return 0.

Step 5

Try every lock that has not yet been broken.

For lock i:

  • Required strength = strength[i]
  • Current factor = x

The waiting time is:

$$\left\lceil \frac{strength[i]}{x} \right\rceil$$

which can be computed as:

$$\frac{strength[i] + x - 1}{x}$$

using integer arithmetic.

Step 6

Break that lock and move to the next state:

$$nextMask = mask ;|; (1<<i)$$

The total cost becomes:

$$wait + dp(nextMask)$$

Step 7

Take the minimum over all possible next locks.

Step 8

Memoize each state so it is computed only once.

Why it Works

For every state, the current growth factor depends only on how many locks have already been broken. The exact order used to reach that state does not matter.

Therefore the future optimal decision depends solely on the current subset of broken locks. The DP explores every possible next lock and chooses the minimum cost continuation. Since every valid lock order corresponds to exactly one path through these states, the DP finds the globally optimal solution.

Python Solution

from typing import List
from functools import lru_cache

class Solution:
    def findMinimumTime(self, strength: List[int], k: int) -> int:
        n = len(strength)
        full_mask = (1 << n) - 1

        @lru_cache(None)
        def dp(mask: int) -> int:
            if mask == full_mask:
                return 0

            broken = mask.bit_count()
            x = 1 + broken * k

            answer = float("inf")

            for i in range(n):
                if not (mask & (1 << i)):
                    wait_time = (strength[i] + x - 1) // x
                    answer = min(
                        answer,
                        wait_time + dp(mask | (1 << i))
                    )

            return answer

        return dp(0)

Implementation Explanation

The recursive function dp(mask) represents the minimum additional time needed after the locks indicated by mask have already been broken.

The number of set bits in mask tells us how many locks have been broken so far. From that value we compute the current growth factor:

x = 1 + broken * k

For every remaining lock, we calculate how many minutes are needed before enough energy accumulates:

wait_time = (strength[i] + x - 1) // x

We then recurse on the state where that lock has been added to the broken set.

Memoization ensures that each subset is solved only once.

Go Solution

func findMinimumTime(strength []int, k int) int {
	n := len(strength)
	fullMask := (1 << n) - 1

	memo := make(map[int]int)

	var bitCount func(int) int
	bitCount = func(x int) int {
		count := 0
		for x > 0 {
			x &= x - 1
			count++
		}
		return count
	}

	var dp func(int) int
	dp = func(mask int) int {
		if mask == fullMask {
			return 0
		}

		if val, ok := memo[mask]; ok {
			return val
		}

		broken := bitCount(mask)
		x := 1 + broken*k

		const INF = int(1e9)
		best := INF

		for i := 0; i < n; i++ {
			if (mask & (1 << i)) == 0 {
				waitTime := (strength[i] + x - 1) / x
				candidate := waitTime + dp(mask|(1<<i))

				if candidate < best {
					best = candidate
				}
			}
		}

		memo[mask] = best
		return best
	}

	return dp(0)
}

Go-Specific Notes

The Go version uses a map for memoization rather than Python's lru_cache.

Since n ≤ 8, the total number of states is at most:

$$2^8 = 256$$

so the memoization map remains very small.

Integer overflow is not a concern because the maximum possible answer is far below Go's int limits.

Worked Examples

Example 1

Input

strength = [3,4,1]
k = 1

Initial state:

State Value
mask 000
broken 0
x 1

Possible first choices:

Lock Wait Time
3 3
4 4
1 1

Optimal choice is lock 1.

After breaking strength 1:

State Value
mask 100
broken 1
x 2

Remaining locks:

Lock Wait Time
3 2
4 2

Choose strength 4.

After breaking strength 4:

State Value
mask 110
broken 2
x 3

Remaining lock:

Lock Wait Time
3 1

Total:

1 + 2 + 1 = 4

Answer:

4

Example 2

Input

strength = [2,5,4]
k = 2

Initial state:

State Value
mask 000
x 1

Choose strength 2 first.

Cost:

ceil(2 / 1) = 2

Now:

State Value
broken 1
x 3

Choose strength 5.

Cost:

ceil(5 / 3) = 2

Now:

State Value
broken 2
x 5

Choose strength 4.

Cost:

ceil(4 / 5) = 1

Total:

2 + 2 + 1 = 5

Answer:

5

Complexity Analysis

Measure Complexity Explanation
Time O(n · 2^n) Each subset examines up to n transitions
Space O(2^n) Memoization stores one value per subset

There are exactly 2^n DP states. For each state we iterate through up to n locks to determine the next choice. Therefore the total running time is O(n · 2^n). The memoization table stores one entry for each subset, giving O(2^n) space usage.

Test Cases

from typing import List

s = Solution()

assert s.findMinimumTime([3, 4, 1], 1) == 4  # example 1
assert s.findMinimumTime([2, 5, 4], 2) == 5  # example 2

assert s.findMinimumTime([1], 1) == 1  # single lock
assert s.findMinimumTime([10], 5) == 10  # single large lock

assert s.findMinimumTime([1, 1], 1) == 2  # identical strengths
assert s.findMinimumTime([2, 2, 2], 2) == 4  # repeated values

assert s.findMinimumTime([100, 1], 1) == 51  # large lock delayed

assert s.findMinimumTime([1, 2, 3, 4], 10) == 4  # large growth factor

assert s.findMinimumTime([1000000], 10) == 1000000  # max strength single lock

assert s.findMinimumTime([8, 6, 4, 2], 1) >= 0  # general sanity test

assert s.findMinimumTime([7, 13, 19, 5, 11, 17, 3, 23], 10) > 0  # max n

Test Summary

Test Why
[3,4,1], k=1 Official example 1
[2,5,4], k=2 Official example 2
[1] Minimum n
[10] Single lock with larger strength
[1,1] Duplicate strengths
[2,2,2] Multiple identical locks
[100,1] Delaying a large lock is beneficial
[1,2,3,4], k=10 Very fast growth factor increase
[1000000] Maximum strength value
[8,6,4,2] General correctness check
Eight-lock test Maximum state-space size

Edge Cases

Single Lock

When there is only one lock, no ordering decisions exist. The answer is simply the time needed to reach its strength with x = 1. A DP implementation can sometimes mishandle this by assuming at least one transition exists. The base case and transition logic correctly handle n = 1.

Extremely Large Strength Values

A lock strength can be as large as 10^6. Simulating energy accumulation minute by minute would be inefficient. The implementation instead computes the waiting time directly using ceiling division:

(strength[i] + x - 1) // x

This keeps every transition constant time.

Large Lock Mixed With Small Locks

Inputs such as:

[100, 1]

can fool greedy strategies. Breaking the large lock first costs 100 minutes, while breaking the small lock first increases the growth factor and reduces the cost of the large lock dramatically. The bitmask DP evaluates all possible orders and therefore always finds the optimal arrangement.

Multiple Locks With Equal Strength

When several locks have the same strength, different orders can lead to the same result. Some implementations may recompute equivalent states repeatedly. Memoization keyed by the subset mask ensures that every state is solved exactly once regardless of how many different paths reach it.