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Minimum Time to Break Locks I

Number: 3649

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

Bob is trapped in a dungeon and must break n locks. Each lock i requires at least strength[i] energy. Bob’s sword starts at 0 energy and with a factor x = 1. Every minute (without breaking a lock) the sword’s energy increases by the current factor x. When Bob breaks a lock, his sword’s energy immediately resets to 0 and the factor x increases by a value k. Bob can choose the order in which he breaks the locks. The goal is to determine the minimum total time (in minutes) required to break all locks.

Key Insights

  • The energy accumulation process for each lock is independent since energy resets after breaking a lock.
  • For a given lock when the sword factor is x, the number of minutes needed is the smallest integer m where m * x >= strength[i] (i.e. m = ceil(strength[i] / x)).
  • The order in which locks are broken affects the current factor x available for each lock, since x starts at 1 and increases by k after each lock is broken.
  • With n up to 8, it is feasible to either brute-force all permutations or use a DP with bitmasking approach to try every possible order.

Space and Time Complexity

Time Complexity: O(n * 2^n) using DP with bitmask (or O(n!) when brute-forcing all permutations, which is acceptable for n ≤ 8).
Space Complexity: O(2^n) for the DP state storage.

Solution

We use a dynamic programming approach with bitmasking. The idea is to let dp[mask] represent the minimum time needed to break the locks corresponding to the bits set in mask. The current sword factor x is determined by how many locks have been already broken (i.e. x = 1 + (number of bits in mask)*k). For each state, we iterate through the remaining locks and compute the minutes needed to break that lock using:
  minutes = ceil(strength[j] / current_factor)
This can be computed exactly as (strength[j] + current_factor - 1) // current_factor in integer math.
We then update the dp state for the new mask including lock j. The answer will be dp[(1 << n) - 1].

Code Solutions

Below are code solutions in Python, JavaScript, C++, and Java.

import math

def minTimeToBreakLocks(strength, k):
    n = len(strength)
    # dp[mask] will hold the minimum time to break locks in the 'mask' configuration.
    dp = [float('inf')] * (1 << n)
    dp[0] = 0
    
    # Iterate over all possible masks.
    for mask in range(1 << n):
        # Count the locks already broken to know current factor: x = 1 + (count * k)
        count = bin(mask).count("1")
        curr_factor = 1 + count * k
        # Try breaking each lock that hasn't been broken.
        for i in range(n):
            if mask & (1 << i) == 0:
                # Calculate minutes required to break lock i with current factor.
                minutes_needed = (strength[i] + curr_factor - 1) // curr_factor
                new_mask = mask | (1 << i)
                dp[new_mask] = min(dp[new_mask], dp[mask] + minutes_needed)
    
    return dp[(1 << n) - 1]

# Example usage:
print(minTimeToBreakLocks([3,4,1], 1))  # Expected output: 4
print(minTimeToBreakLocks([2,5,4], 2))  # Expected output: 5
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