Problem Description

You have a combination lock with 4 circular wheels, where each wheel can display digits from '0' to '9'. The wheels can rotate in both directions and wrap around - meaning you can turn '9' to '0' or '0' to '9'. In each move, you can turn exactly one wheel by one position (either forward or backward).

The lock starts at the initial position '0000'. You need to reach a given target combination to unlock it.

However, there are certain combinations called deadends that will cause the lock to jam. If the lock ever displays any of these deadend combinations, it becomes stuck and cannot be turned further.

Your task is to find the minimum number of moves required to change the lock from '0000' to the target combination without ever landing on a deadend. If it's impossible to reach the target (either because the starting position is a deadend or all paths to the target go through deadends), return -1.

For example:

• From '0000', you could move to '1000', '9000', '0100', '0900', '0010', '0090', '0001', or '0009' in one move
• If target = '0202' and there are no deadends, you would need at least 4 moves (turning the second wheel twice and the fourth wheel twice)
• If '0000' itself is in the deadends list, it's impossible to start, so return -1

Flowchart Walkthrough

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

Is it a graph?

• Yes: This problem can be modeled as a graph where each lock combination (like '0000', '0001', etc.) is a node, and edges connect combinations that differ by one wheel turn. We need to find the shortest path from '0000' to the target.

Is it a tree?

• No: The graph is not a tree because multiple paths can lead to the same combination, and there can be cycles (you can return to a previous combination).

Is the problem related to directed acyclic graphs?

• No: The graph has cycles (you can go from '0000' to '1000' and back to '0000').

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of moves (shortest path) from the starting combination '0000' to the target combination.

Is the graph Weighted?

• No: Each move (edge) has the same cost of 1. All edges have equal weight since each represents exactly one wheel turn.

BFS

• Yes: Since we have an unweighted graph and need the shortest path, BFS is the optimal choice.

Conclusion: The flowchart correctly leads us to use BFS (Breadth-First Search) for this problem. BFS is ideal here because:

1. We're exploring an implicit graph of lock combinations
2. All moves have equal cost (unweighted edges)
3. We need the minimum number of moves (shortest path)
4. BFS guarantees finding the shortest path in unweighted graphs

Intuition

Think of this problem as navigating through a maze where each position in the maze is a 4-digit combination. You start at '0000' and want to reach your target combination, but some positions in the maze are blocked (the deadends).

The key insight is that from any combination, you can move to exactly 8 neighboring combinations - by turning each of the 4 wheels either forward or backward by one position. For example, from '0000', you can reach '1000', '9000', '0100', '0900', '0010', '0090', '0001', or '0009'.

Since each move changes exactly one wheel by one position, every move has the same "cost" of 1. This transforms our problem into finding the shortest path in an unweighted graph, where BFS naturally gives us the optimal solution.

Why BFS works perfectly here:

1. Level-by-level exploration: BFS explores all combinations reachable in 1 move, then 2 moves, then 3 moves, and so on. The first time we encounter the target, we've found it using the minimum number of moves.

2. Avoiding revisits: We need to track visited combinations to avoid going in circles. Once we've reached a combination with n moves, there's no point reaching it again with more moves.

3. Handling deadends: We treat deadends as pre-visited nodes - we simply never explore them. If '0000' itself is a deadend, we can't even start, so we return -1.

The algorithm essentially performs a breadth-first search starting from '0000', exploring all valid neighboring combinations level by level, until we either find the target (return the number of moves) or exhaust all possibilities (return -1).

Learn more about Breadth-First Search patterns.

Solution Approach

The implementation uses BFS with a queue to explore combinations level by level. Let's walk through the key components:

1. Generating Neighbors The next(s) function generates all 8 possible combinations reachable from the current state:

• For each of the 4 wheels (positions 0-3)
• Try turning it backward: '0' → '9', or decrement by 1
• Try turning it forward: '9' → '0', or increment by 1
• This gives us 4 wheels × 2 directions = 8 neighbors

2. Initial Setup

• Check if target is already '0000' - return 0 immediately
• Convert deadends to a set s for O(1) lookup
• Check if '0000' is a deadend - return -1 if we can't start
• Initialize queue with '0000' and add it to visited set

3. BFS Algorithm

while q:
    ans += 1  # Increment level/distance
    for _ in range(len(q)):  # Process all nodes at current level
        p = q.popleft()
        for t in next(p):  # Check all 8 neighbors
            if t == target:
                return ans  # Found target at this level
            if t not in s:  # Not visited and not a deadend
                q.append(t)
                s.add(t)  # Mark as visited

Key Data Structures:

• Queue (deque): Enables efficient FIFO operations for BFS
• Set (s): Stores both deadends and visited combinations for O(1) lookup
• String manipulation: Each combination is stored as a 4-character string

Why This Works:

1. The level-by-level processing ensures we find the shortest path
2. The visited set prevents revisiting combinations (avoiding infinite loops)
3. Deadends are pre-added to the visited set, effectively blocking those paths
4. The first time we encounter the target, we've found the minimum number of moves

Time Complexity: O(10^4 × 8) = O(80,000) in worst case, as there are 10^4 possible combinations and we might need to explore most of them, with 8 operations per combination.

Space Complexity: O(10^4) for storing visited combinations in the set and queue.

Example Walkthrough

Let's walk through a concrete example where target = "0202" and deadends = ["0201", "0101", "0102", "1212", "2002"].

Initial State:

• Start: "0000"
• Target: "0202"
• Deadends set: {"0201", "0101", "0102", "1212", "2002"}

Level 0 (0 moves):

• Queue: ["0000"]
• Check if "0000" is target? No, continue.

Level 1 (1 move):

• Process "0000" and generate 8 neighbors:
  • "1000" ✓ (valid, add to queue)
  • "9000" ✓ (valid, add to queue)
  • "0100" ✓ (valid, add to queue)
  • "0900" ✓ (valid, add to queue)
  • "0010" ✓ (valid, add to queue)
  • "0090" ✓ (valid, add to queue)
  • "0001" ✓ (valid, add to queue)
  • "0009" ✓ (valid, add to queue)
• None match target, continue to next level.

Level 2 (2 moves):

• From "0100", we can reach:
  • "0200" ✓ (valid, not a deadend, add to queue)
  • "0101" ✗ (deadend, skip)
  • Other neighbors...
• From other nodes in level 1, generate their neighbors
• Still no target found, continue.

Level 3 (3 moves):

• From "0200", we can reach:
  • "0201" ✗ (deadend, skip)
  • "0300" ✓ (valid)
  • "0100" ✗ (already visited)
  • Other neighbors...
• Continue processing other nodes at this level.

Level 4 (4 moves):

• Eventually, from some combination (like "0203"), we generate:
  • "0202" → Target found!
• Return 4 moves.

The BFS ensures we explore all combinations reachable in 1 move, then 2 moves, then 3 moves, and so on. When we first encounter "0202", we know we've found it using the minimum number of moves (4 in this case). The deadends "0201", "0101", and "0102" blocked certain paths, but BFS found an alternative route that avoided them.

Solution Implementation

Time and Space Complexity

Time Complexity: O(10^4 * 4 * 2) = O(10^4)

The algorithm uses BFS to explore all possible lock combinations. In the worst case:

• There are 10^4 possible combinations (4 digits, each ranging from 0-9)
• For each combination visited, the next() function generates neighbors by:
  • Iterating through 4 positions
  • For each position, creating 2 new combinations (increment and decrement)
  • Each string operation inside next() takes O(4) time for joining
• So processing each node takes O(4 * 2 * 4) = O(32) = O(1) time
• Total time: O(10^4 * 1) = O(10^4)

Space Complexity: O(10^4)

The space is dominated by:

• The set s which stores visited combinations and deadends: up to O(10^4) entries in worst case
• The queue q which can contain up to O(10^4) combinations in worst case
• The next() function creates temporary lists and strings, but this is O(1) space per call
• Total space: O(10^4)

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

Common Pitfalls

1. Forgetting to Check if Starting Position is a Deadend

One of the most common mistakes is not checking whether '0000' itself is in the deadends list before starting the BFS. If the starting position is blocked, the lock is immediately stuck and cannot make any moves.

Incorrect approach:

# Missing the initial deadend check
visited = set(deadends)
queue = deque(['0000'])
visited.add('0000')  # This adds '0000' even if it's a deadend!

Correct approach:

visited = set(deadends)
if '0000' in visited:  # Check BEFORE adding to queue
    return -1
queue = deque(['0000'])
visited.add('0000')

2. Inefficient Level Tracking in BFS

Another pitfall is incrementing the step counter for each node processed rather than for each level. This leads to incorrect distance calculations.

Incorrect approach:

while queue:
    current = queue.popleft()
    steps += 1  # Wrong! This increments for every node
    for next_state in get_next_states(current):
        if next_state == target:
            return steps

Correct approach:

while queue:
    steps += 1  # Increment once per level
    for _ in range(len(queue)):  # Process all nodes at current level
        current = queue.popleft()
        for next_state in get_next_states(current):
            if next_state == target:
                return steps

3. String Mutation Issues When Generating Neighbors

Python strings are immutable, so attempting to modify them directly won't work. Additionally, forgetting to restore the original digit when generating multiple neighbors can lead to incorrect states.

Incorrect approach:

def get_next_states(state):
    next_states = []
    for i in range(4):
        state[i] = str((int(state[i]) + 1) % 10)  # Error: strings are immutable
        next_states.append(state)
        state[i] = str((int(state[i]) - 1) % 10)  # Wrong calculation
        next_states.append(state)

Correct approach:

def get_next_states(state):
    next_states = []
    state_list = list(state)  # Convert to mutable list
  
    for i in range(4):
        original = state_list[i]  # Save original
      
        # Generate both neighbors
        state_list[i] = '9' if original == '0' else str(int(original) - 1)
        next_states.append(''.join(state_list))
      
        state_list[i] = '0' if original == '9' else str(int(original) + 1)
        next_states.append(''.join(state_list))
      
        state_list[i] = original  # Restore for next iteration

4. Not Handling the Edge Case Where Target is '0000'

If the target is the starting position itself, the answer should be 0, not going through the entire BFS process.

Solution: Always check this edge case at the beginning:

if target == '0000':
    return 0