LeetCode 3385 - Minimum Time to Break Locks II

Difficulty: 🔴 Hard
Topics: Array, Breadth-First Search, Graph Theory

Solution

Problem Understanding

In this problem, Bob must break a collection of locks, where each lock requires a certain minimum amount of energy before it can be destroyed. The input array strength represents these requirements. If strength[i] = s, then Bob must wait until the sword's energy reaches at least s before breaking that lock.

The sword behaves in a very specific way:

Initially, the sword's energy is 0
Initially, the energy growth factor X = 1
Every minute, the sword gains X energy
After breaking a lock:
energy resets to 0
X increases by 1

The important detail is that the growth factor permanently increases after every lock break. That means the order in which we break locks directly affects the total time.

Suppose Bob has already broken k locks. Then the current factor is:

$$X = k + 1$$

If the next lock requires energy s, the waiting time needed is:

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

because the sword gains X energy per minute.

The goal is to choose the optimal order of breaking locks so that the total waiting time is minimized.

The constraints are very important:

1 <= n <= 80
1 <= strength[i] <= 10^6

A naive brute-force search over all possible orders would require checking n! permutations, which becomes impossible even for moderate n.

The problem guarantees:

Every strength value is positive
The number of locks is relatively small
We only need the minimum total time

An important observation is that after every lock break, the factor increases regardless of which lock was chosen. Therefore, each position in the order corresponds to a specific factor value.

Several edge cases are worth considering:

A single lock
Multiple identical strengths
Very large strength values
Locks already sorted in ascending or descending order
Cases where a greedy local choice might fail

These cases can expose incorrect assumptions about optimal ordering.

Approaches

Brute Force Approach

The brute-force solution tries every possible order of breaking locks.

For each permutation:

Start with X = 1
For every lock in the chosen order:

compute required waiting time:

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

add it to the total
increment X

After evaluating all permutations, return the minimum total time.

This approach is correct because it explicitly checks every possible valid ordering, so the optimal one must appear somewhere in the search space.

However, the complexity is catastrophic:

$$O(n! \cdot n)$$

With n = 80, this is completely infeasible.

Key Insight

The critical observation is that larger factors should generally be paired with larger strengths.

Suppose two locks a and b are broken with factors x and x+1.

We compare:

break a first, then b

versus

break b first, then a

The total cost becomes:

$$\left\lceil \frac{a}{x} \right\rceil + \left\lceil \frac{b}{x+1} \right\rceil$$

versus

$$\left\lceil \frac{b}{x} \right\rceil + \left\lceil \frac{a}{x+1} \right\rceil$$

Intuitively, larger strengths benefit more from larger divisors. Therefore, we should use smaller factors on smaller strengths and larger factors on larger strengths.

This leads to a simple greedy strategy:

Sort the strengths in ascending order
Break locks from smallest to largest

Once sorted, the i-th lock is broken with factor:

$$X = i + 1$$

The total answer becomes:

$$\sum_{i=0}^{n-1} \left\lceil \frac{\text{strength}[i]}{i+1} \right\rceil$$

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n! · n)	O(n)	Tries every possible order
Optimal Greedy	O(n log n)	O(1) or O(n)	Sort ascending and pair small strengths with small factors

Algorithm Walkthrough

Sort the strength array in ascending order.

This ensures that weaker locks are handled earlier when the sword factor is still small, while stronger locks benefit from larger factors later. 2. Initialize total_time = 0.

This variable stores the cumulative minutes required to break all locks. 3. Iterate through the sorted array using index i.

Since Bob starts with factor 1 and gains one factor after every lock, the current factor is:

$$X = i + 1$$ 4. Compute the waiting time for the current lock.

The sword gains X energy per minute, so the number of minutes required is:

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

Integer ceiling division can be implemented as:

(s + X - 1) // X

Add this waiting time to total_time.
Continue until all locks are processed.
Return total_time.

Why it works

The greedy ordering is optimal because larger factors reduce waiting time more significantly for larger strengths. Sorting strengths in ascending order guarantees that larger locks receive larger multipliers.

This is a classic exchange argument. If two adjacent locks are out of order, swapping them cannot increase the total cost when the smaller lock comes first. Repeatedly applying this argument transforms any optimal ordering into sorted order.

Python Solution

from typing import List

class Solution:
    def findMinimumTime(self, strength: List[int]) -> int:
        strength.sort()

        total_time = 0

        for index, value in enumerate(strength):
            factor = index + 1

            # Ceiling division
            minutes = (value + factor - 1) // factor

            total_time += minutes

        return total_time

The implementation begins by sorting the array in ascending order. This directly applies the greedy insight that smaller locks should use smaller sword factors.

The loop processes each lock in order. The variable factor represents the current energy growth rate after breaking previous locks.

For each lock, we compute how many minutes are needed for the sword energy to reach the required strength. Since energy increases linearly by factor every minute, we use ceiling division.

The result is accumulated into total_time, which is returned at the end.

The implementation is concise because the mathematical observation eliminates the need for dynamic programming or graph traversal.

Go Solution

package main

import (
	"sort"
)

func findMinimumTime(strength []int) int {
	sort.Ints(strength)

	totalTime := 0

	for i, value := range strength {
		factor := i + 1

		// Ceiling division
		minutes := (value + factor - 1) / factor

		totalTime += minutes
	}

	return totalTime
}

The Go implementation follows the exact same logic as the Python version.

A few Go-specific details are worth mentioning:

sort.Ints is used for in-place sorting
Integer division in Go automatically truncates toward zero, which makes the ceiling division formula work correctly
Since the maximum possible answer easily fits inside a 32-bit integer, standard int is sufficient

Worked Examples

Example 1

Input:

strength = [3,4,1]

After sorting:

[1,3,4]

Step	Lock Strength	Factor X	Minutes Needed	Total Time
1	1	1	ceil(1/1) = 1	1
2	3	2	ceil(3/2) = 2	3
3	4	3	ceil(4/3) = 2	5

Final answer:

Example 2

Input:

strength = [2,5,4]

After sorting:

[2,4,5]

Step	Lock Strength	Factor X	Minutes Needed	Total Time
1	2	1	ceil(2/1) = 2	2
2	4	2	ceil(4/2) = 2	4
3	5	3	ceil(5/3) = 2	6

Final answer:

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates the runtime
Space	O(1) or O(n)	Depends on sorting implementation

The loop itself is linear, requiring only one pass through the array. The dominant operation is sorting, which costs O(n log n).

No additional data structures proportional to input size are required beyond the sorting implementation.

Test Cases

from typing import List

class Solution:
    def findMinimumTime(self, strength: List[int]) -> int:
        strength.sort()

        total_time = 0

        for index, value in enumerate(strength):
            factor = index + 1
            total_time += (value + factor - 1) // factor

        return total_time

sol = Solution()

assert sol.findMinimumTime([3, 4, 1]) == 5  # Basic unsorted example
assert sol.findMinimumTime([2, 5, 4]) == 6  # Example from statement
assert sol.findMinimumTime([1]) == 1  # Single lock
assert sol.findMinimumTime([10]) == 10  # Single large lock
assert sol.findMinimumTime([1, 1, 1]) == 3  # All identical small values
assert sol.findMinimumTime([5, 5, 5]) == 10  # Identical medium values
assert sol.findMinimumTime([1, 2, 3, 4]) == 5  # Already sorted
assert sol.findMinimumTime([4, 3, 2, 1]) == 5  # Reverse sorted
assert sol.findMinimumTime([1000000]) == 1000000  # Maximum strength single value
assert sol.findMinimumTime([8, 1, 2, 9]) == 9  # Mixed ordering
assert sol.findMinimumTime([7, 7, 1, 1]) == 10  # Duplicate groups

Test Summary

Test	Why
`[3,4,1]`	Basic unsorted input
`[2,5,4]`	Matches problem example
`[1]`	Minimum input size
`[10]`	Single large lock
`[1,1,1]`	All equal small values
`[5,5,5]`	All equal medium values
`[1,2,3,4]`	Already sorted input
`[4,3,2,1]`	Reverse sorted input
`[1000000]`	Maximum strength value
`[8,1,2,9]`	Random mixed ordering
`[7,7,1,1]`	Duplicate clusters

Edge Cases

Single Lock

If there is only one lock, the factor always remains 1. The answer is simply the strength itself.

A buggy implementation might incorrectly increment the factor before processing the first lock. This implementation avoids that by using:

factor = index + 1

so the first lock always uses factor 1.

Duplicate Strength Values

When multiple locks have the same strength, the ordering between them does not matter mathematically. However, some implementations may accidentally rely on unique values or unstable indexing logic.

Sorting handles duplicates naturally, and each duplicate simply receives progressively larger factors.

Very Large Strength Values

Strength values can reach 10^6. A naive simulation that increments energy minute by minute would be far too slow.

This implementation never simulates energy accumulation directly. Instead, it computes the required waiting time using ceiling division in constant time:

(value + factor - 1) // factor

This guarantees efficiency even for the largest inputs.