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Minimum Time to Break Locks II

Number: 3693

Difficulty: Hard

Paid? Yes

Companies: N/A

Problem Description

Bob is trapped in a dungeon with n locks. Each lock i requires a certain amount of energy given by strength[i] to break. Bob’s sword starts with 0 energy and a base multiplier X = 1. Every minute, the sword’s energy increases by the current factor X. In order to break a lock, the sword’s current energy must reach at least the required strength. When Bob breaks a lock, the sword’s energy resets to 0 and the factor X increases by 1. The goal is to determine the minimum total time (in minutes) required for Bob to break all locks. Note that Bob is allowed to choose the order in which he breaks the locks.

Key Insights

  • When attempting a lock with current factor f, Bob must wait t minutes such that t * f >= required strength. This “waiting time” can be computed as ceil(strength / f).
  • Since after breaking a lock the factor increases (from 1 up to n across locks), the order in which locks are broken changes the waiting time per lock.
  • You can view the problem as assigning each lock a unique “slot” (i.e. a factor from 1 to n). The cost of assigning lock i to slot j is ceil(strength[i] / j).
  • The challenge reduces to the classic assignment problem: assign each lock a multiplier slot so that the total sum of waiting times is minimized.
  • The Hungarian algorithm (or any minimum-cost bipartite matching algorithm) is well-suited to solve this assignment problem efficiently given that n ≤ 80.

Space and Time Complexity

Time Complexity: O(n^3), due to the Hungarian algorithm over an n×n cost matrix. Space Complexity: O(n^2), to store the cost matrix.

Solution

We create an n×n cost matrix where the entry in row i and column j is the cost (number of minutes) needed to break lock i when using factor (j+1) (since factors start at 1). The cost can be computed as:   cost[i][j] = (strength[i] + j) // (j+1) After forming the matrix, we run the Hungarian algorithm to find the assignment (a permutation of locks to multiplier slots) that minimizes the total sum of costs. The answer is that minimum sum.

Data structures used:

  • A 2-dimensional array (matrix) to represent the cost between each lock and each slot.
  • Arrays for maintaining labels, matching indices, and slack values in the Hungarian algorithm.

Algorithmic approach:

  1. Build the cost matrix based on the formula above.
  2. Use the Hungarian algorithm to solve the minimum cost assignment.
  3. Return the total minimum cost.

Important points:

  • Converting the wait time to a cost via ceiling division is the key “trick” to modeling the lock-breaking process.
  • Recognizing the assignment formulation allows the use of a well-known algorithm rather than attempting a combinatorial permutation search.

Code Solutions

def min_time_to_break_locks(strength):
    import math
    n = len(strength)
    # Build the cost matrix: cost[i][j] when lock i is attempted with factor (j+1)
    cost = [[(strength[i] + j) // (j+1) for j in range(n)] for i in range(n)]
    
    # Hungarian algorithm (for minimization).
    N = n
    u = [0] * (N+1)
    v = [0] * (N+1)
    p = [0] * (N+1)
    way = [0] * (N+1)
    for i in range(1, N+1):
        p[0] = i
        minv = [float('inf')] * (N+1)
        used = [False] * (N+1)
        j0 = 0
        while True:
            used[j0] = True
            i0 = p[j0]
            delta = float('inf')
            j1 = 0
            for j in range(1, N+1):
                if not used[j]:
                    cur = cost[i0-1][j-1] - u[i0] - v[j]
                    if cur < minv[j]:
                        minv[j] = cur
                        way[j] = j0
                    if minv[j] < delta:
                        delta = minv[j]
                        j1 = j
            for j in range(N+1):
                if used[j]:
                    u[p[j]] += delta
                    v[j] -= delta
                else:
                    minv[j] -= delta
            j0 = j1
            if p[j0] == 0:
                break
        # Augmenting path update
        while True:
            j1 = way[j0]
            p[j0] = p[j1]
            j0 = j1
            if j0 == 0:
                break
    # The matching cost will be -v[0]
    return -v[0]

# Example test cases
print(min_time_to_break_locks([3,4,1]))  # Expected output 4
print(min_time_to_break_locks([2,5,4]))  # Expected output 6

# The function min_time_to_break_locks returns the minimum total waiting time (in minutes).
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