Problem Description

Bob needs to break n locks to escape from a dungeon. Each lock requires a certain amount of energy to break, given in an array strength where strength[i] represents the energy required to break the i-th lock.

Bob has a sword that works with the following mechanics:

1. Initial state: The sword starts with 0 energy and a factor x = 1

2. Energy accumulation: Every minute that passes, the sword's energy increases by the current factor x. For example:

   • After 1 minute: energy = x
   • After 2 minutes: energy = 2x
   • After t minutes: energy = t * x

3. Breaking locks: To break the i-th lock, the sword's energy must be at least strength[i]. Bob can choose which lock to break in any order.

4. Reset and upgrade: After breaking any lock:

   • The sword's energy resets to 0
   • The factor x increases by K (so x becomes x + K)

The goal is to find the minimum total time (in minutes) needed to break all n locks.

For example, if Bob breaks locks in some order, and the factor starts at 1:

• For the first lock: factor is 1, time needed is ⌈strength[first_lock] / 1⌉ minutes
• For the second lock: factor is 1 + K, time needed is ⌈strength[second_lock] / (1 + K)⌉ minutes
• For the third lock: factor is 1 + 2K, time needed is ⌈strength[third_lock] / (1 + 2K)⌉ minutes
• And so on...

The challenge is to determine the optimal order to break the locks to minimize the total time.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: This problem doesn't explicitly involve nodes and edges. However, we can model it as a state-space problem where each state represents which locks have been broken.

Does the problem have small constraints?

• Yes: With n locks, we need to explore all possible orderings of breaking the locks. The solution uses bit manipulation with states up to (1 << len(strength)) - 1, indicating we're dealing with all 2^n possible subsets of locks. This is typically feasible only for small n (usually n ≤ 20).

DFS/backtracking?

• Yes: We need to explore all possible permutations of breaking the locks to find the optimal order. The solution uses DFS with memoization (@cache) to explore different sequences of breaking locks.

The DFS approach is evident in the solution:

• dfs(i) represents the minimum time needed when the state i (a bitmask) indicates which locks have already been broken
• For each unbroken lock (checked with i >> j & 1 ^ 1), we recursively try breaking it next
• We explore all possible next locks to break and choose the minimum time
• The base case is when all locks are broken (i == (1 << len(strength)) - 1)

Conclusion: The flowchart correctly leads us to use DFS/backtracking. The problem requires exploring all possible orderings of breaking locks (permutations), which is a classic application of backtracking with state space exploration. The small constraints (implied by the bit manipulation approach) make this brute force approach feasible.

Intuition

The key insight is that the order in which we break the locks significantly affects the total time needed. Since the factor x increases by K after each lock is broken, locks broken later will benefit from a higher factor, requiring less time.

Let's think about this step by step:

1. Why order matters: If we have two locks with strengths [10, 20] and K = 2:

   • Breaking lock 1 first, then lock 2: ⌈10/1⌉ + ⌈20/3⌉ = 10 + 7 = 17 minutes
   • Breaking lock 2 first, then lock 1: ⌈20/1⌉ + ⌈10/3⌉ = 20 + 4 = 24 minutes

   The order clearly makes a difference!

2. The state space: At any point, we need to know which locks have been broken and which remain. This naturally leads to thinking about subsets of locks. With n locks, there are 2^n possible states (each lock is either broken or not).

3. Optimal substructure: Once we've broken some locks, the minimum time to break the remaining locks depends only on:

   • Which locks remain (not the order we broke previous locks)
   • The current factor x (which is determined by how many locks we've broken: x = 1 + count * K)

4. Why DFS with memoization: We need to try all possible next locks to break and choose the best option. This is a classic recursive structure. However, we might reach the same state (same set of broken locks) through different paths, so memoization helps avoid recalculating.

5. Bitmask representation: Since we need to track subsets of locks and n is small, using a bitmask is perfect. Each bit represents whether a lock is broken (1) or not (0). For example, 101 means locks 0 and 2 are broken, lock 1 remains.

The formula (s + x - 1) // x calculates the ceiling division - the minimum minutes needed to accumulate at least s energy when gaining x energy per minute. This is equivalent to math.ceil(s / x).

By exploring all possible orderings through DFS and caching intermediate results, we ensure we find the globally optimal solution without redundant calculations.

Learn more about Depth-First Search, Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The solution implements a DFS with memoization using bit manipulation to track states. Let's break down the implementation:

1. State Representation with Bitmask:

• We use an integer i as a bitmask where each bit represents whether a lock is broken
• i = 0 means no locks are broken (initial state)
• i = (1 << len(strength)) - 1 means all locks are broken (terminal state)
• For example, with 3 locks: i = 5 (binary 101) means locks 0 and 2 are broken

2. The DFS Function:

@cache
def dfs(i: int) -> int:

• dfs(i) returns the minimum time needed to break all remaining locks when state is i
• The @cache decorator memoizes results to avoid recalculating same states

3. Base Case:

if i == (1 << len(strength)) - 1:
    return 0

• When all bits are set (all locks broken), no more time is needed

4. Calculate Current Factor:

cnt = i.bit_count()
x = 1 + cnt * K

• bit_count() returns the number of 1s in the bitmask (number of broken locks)
• The current factor is 1 + (number of broken locks) * K

5. Try Breaking Each Remaining Lock:

for j, s in enumerate(strength):
    if i >> j & 1 ^ 1:
        ans = min(ans, dfs(i | 1 << j) + (s + x - 1) // x)

• i >> j & 1 ^ 1 checks if the j-th bit is 0 (lock j is not broken)
  • i >> j shifts right by j positions
  • & 1 gets the j-th bit
  • ^ 1 flips the bit (so we get True if bit was 0)
• i | 1 << j sets the j-th bit to 1 (marking lock j as broken)
• (s + x - 1) // x calculates ceiling division - time needed to accumulate at least s energy with factor x

6. Return Minimum:

• We recursively calculate the time for each possible next lock and return the minimum total time

Time Complexity: O(n * 2^n) where n is the number of locks. We have 2^n states and for each state, we check n locks.

Space Complexity: O(2^n) for the memoization cache.

The algorithm systematically explores all possible orderings by trying each unbroken lock as the next one to break, leveraging memoization to ensure each state is computed only once.

Example Walkthrough

Let's walk through a small example with strength = [3, 5] and K = 2.

We need to find the minimum time to break both locks. Let's explore both possible orders:

Order 1: Break lock 0 first, then lock 1

• Initial state: bitmask = 00 (binary), no locks broken
• Factor x = 1 + 0*2 = 1
• Break lock 0 (strength = 3): Time = ⌈3/1⌉ = 3 minutes
• State after: bitmask = 01 (lock 0 broken)
• Factor x = 1 + 1*2 = 3
• Break lock 1 (strength = 5): Time = ⌈5/3⌉ = 2 minutes
• Total time: 3 + 2 = 5 minutes

Order 2: Break lock 1 first, then lock 0

• Initial state: bitmask = 00 (binary), no locks broken
• Factor x = 1 + 0*2 = 1
• Break lock 1 (strength = 5): Time = ⌈5/1⌉ = 5 minutes
• State after: bitmask = 10 (lock 1 broken)
• Factor x = 1 + 1*2 = 3
• Break lock 0 (strength = 3): Time = ⌈3/3⌉ = 1 minute
• Total time: 5 + 1 = 6 minutes

DFS Execution Trace:

dfs(00) - Need to break all locks from initial state
  Try lock 0: dfs(01) + 3
    dfs(01) - Only lock 1 remains
      Try lock 1: dfs(11) + 2
        dfs(11) = 0 (base case, all locks broken)
      Return 2
  Try lock 1: dfs(10) + 5
    dfs(10) - Only lock 0 remains
      Try lock 0: dfs(11) + 1
        dfs(11) = 0 (cached result)
      Return 1
  Return min(3+2, 5+1) = min(5, 6) = 5

The optimal solution is to break lock 0 first, then lock 1, taking 5 minutes total. The algorithm explores both orderings and picks the minimum.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * 2^n) where n is the length of the strength array.

The algorithm uses dynamic programming with bit masking to explore all possible orderings of locks. The state space has 2^n possible states (each subset of locks that have been broken), and for each state, we iterate through all n locks to find the next lock to break. The memoization ensures each state is computed only once, giving us O(n * 2^n) time complexity.

Space Complexity: O(2^n) where n is the length of the strength array.

The space is dominated by the memoization cache which stores the result for each possible state (bitmask). There are 2^n possible states, and each state stores a single integer value. Additionally, the recursion stack can go up to depth n in the worst case, but this O(n) is dominated by the cache size of O(2^n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Ceiling Division Formula

Pitfall: Using math.ceil(s / x) instead of (s + x - 1) // x can cause floating-point precision errors, especially when dealing with large numbers.

# Incorrect - can have precision issues
import math
time_needed = math.ceil(strength[j] / x)

# Correct - integer arithmetic only
time_needed = (strength[j] + x - 1) // x

Why it matters: Floating-point division can introduce rounding errors that might give incorrect results for edge cases. Integer-only arithmetic guarantees exact results.

2. Wrong Bit Manipulation for Checking Unbroken Locks

Pitfall: Incorrectly checking if a lock is broken or using wrong operator precedence.

# Common mistakes:
if i & (1 << j) == 0:  # Correct but verbose
if i >> j & 1 ^ 1:      # Works but confusing precedence
if not (i >> j & 1):    # Clearer alternative

# Wrong implementations:
if i >> j & 1:          # This checks if lock IS broken (opposite logic)
if i & 1 << j ^ 1:      # Wrong precedence - XOR applies to wrong part

Solution: Use parentheses for clarity or more explicit conditions:

# Clear and correct alternatives:
if (mask >> task_idx) & 1 == 0:  # Explicitly check if bit is 0
if not ((mask >> task_idx) & 1): # Using not operator
if mask & (1 << task_idx) == 0:  # Check without shifting

3. Forgetting to Handle Empty Input or Edge Cases

Pitfall: Not considering edge cases like empty strength array or K = 0.

class Solution:
    def findMinimumTime(self, strength: List[int], K: int) -> int:
        # Add edge case handling
        if not strength:
            return 0
        if all(s == 0 for s in strength):
            return 0
          
        # Rest of the solution...

4. Integer Overflow in Bit Operations

Pitfall: With many locks (n > 20), the bitmask approach might face memory issues since we need 2^n states.

# Check constraints first
if len(strength) > 20:
    # Consider alternative approach for large n
    # The bitmask DP won't be feasible
    pass

5. Not Using @cache or Implementing Manual Memoization Incorrectly

Pitfall: Forgetting the @cache decorator or implementing manual memoization with mutable default arguments.

# Wrong - mutable default argument
def dfs(mask, memo={}):  # Dictionary is shared across calls!
    if mask in memo:
        return memo[mask]
    # ...
  
# Correct approaches:
# 1. Using @cache (simplest)
@cache
def dfs(mask):
    # ...

# 2. Manual memoization (if needed)
def findMinimumTime(self, strength, K):
    memo = {}  # Create fresh dictionary for each call
    def dfs(mask):
        if mask in memo:
            return memo[mask]
        # ... calculate result ...
        memo[mask] = result
        return result

6. Off-by-One Error in Terminal State Check

Pitfall: Incorrectly calculating the "all locks broken" state.

# Wrong implementations:
if mask == (1 << len(strength)):      # One too large
if mask == (1 << len(strength) - 1):  # Operator precedence issue

# Correct:
if mask == (1 << len(strength)) - 1:  # All n bits set to 1
# For n=3: (1 << 3) - 1 = 8 - 1 = 7 = 0b111

Best Practice Solution:

class Solution:
    def findMinimumTime(self, strength: List[int], K: int) -> int:
        from functools import cache
      
        n = len(strength)
        if n == 0:
            return 0
          
        ALL_COMPLETE = (1 << n) - 1  # Define constant for clarity
      
        @cache
        def dfs(mask: int) -> int:
            if mask == ALL_COMPLETE:
                return 0
          
            completed = bin(mask).count('1')  # Alternative to bit_count()
            power = 1 + completed * K
          
            min_time = float('inf')
            for i in range(n):
                if not (mask & (1 << i)):  # Clear check for unset bit
                    time_needed = -(-strength[i] // power)  # Alternative ceiling division
                    min_time = min(min_time, dfs(mask | (1 << i)) + time_needed)
          
            return min_time
      
        return dfs(0)