How do you solve Minimum Time to Break Locks I on LeetCode?

Minimum Time to Break Locks I (LeetCode #3376) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^2) time complexity and O(1) space complexity. Key insight: Sorting the locks helps in minimizing the time taken.

What is the brute force solution for Minimum Time to Break Locks I?

The brute force approach for Minimum Time to Break Locks I is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach involves trying every possible order of breaking the locks. Since there are n! permutations of the locks, we can calculate the time taken for each permutation and find the minimum time required. The optimal Optimal Solution approach improves this to O(n^2).

What are the interview tips for Minimum Time to Break Locks I?

Explain your thought process clearly. Discuss edge cases, such as the smallest and largest inputs. Practice time complexity analysis for different approaches.

What are common mistakes when solving Minimum Time to Break Locks I?

Not considering all permutations in brute force. Failing to reset energy and factor correctly.

#3376

Minimum Time to Break Locks I

Medium

ArrayDynamic ProgrammingBacktrackingBit ManipulationBreadth-First SearchBitmaskBacktrackingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n^2)
Space	O(n)	O(1)

💡

Intuition

Time O(n^2)Space O(1)

Instead of checking all permutations, we can use a backtracking approach to explore breaking locks in a systematic way. We can keep track of the current energy and factor, and only move forward when we can break a lock.

⚙️

Algorithm

3 steps

1Step 1: Sort the locks based on their strength.
2Step 2: Use backtracking to try breaking locks in order, keeping track of time, energy, and factor.
3Step 3: Return the minimum time found.

solution.py12 lines

1def minTimeToBreakLocks(strength, k):
2    strength.sort()
3    return backtrack(strength, 0, 0, 1)
4
5def backtrack(strength, index, time, factor):
6    if index == len(strength):
7        return time
8    energy = 0
9    while energy < strength[index]:
10        time += 1
11        energy += factor
12    return backtrack(strength, index + 1, time, factor + k)

ℹ

Complexity note: The time complexity is O(n^2) in the worst case due to the while loop for each lock. The space complexity is O(1) since we are using constant space for variables.

1Sorting the locks helps in minimizing the time taken.
2Using backtracking allows us to systematically explore options.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.