752. Open The Lock

Medium

You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot.

The lock initially starts at '0000', a string representing the state of the 4 wheels.

You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.

Hints:

We can think of this problem as a shortest path problem on a graph: there are `10000` nodes (strings `'0000'` to `'9999'`), and there is an edge between two nodes if they differ in one digit, that digit differs by 1 (wrapping around, so `'0'` and `'9'` differ by 1), and if *both* nodes are not in `deadends`.

SolvingApproachs:

Approach #1: Breadth-First Search [Accepted]

Intuition

We can think of this problem as a shortest path problem on a graph: there are 10000 nodes (strings '0000' to '9999'), and there is an edge between two nodes if they differ in one digit, that digit differs by 1 (wrapping around, so '0' and '9' differ by 1), and if both nodes are not in deadends.

Algorithm

To solve a shortest path problem, we use a breadth-first search. The basic structure uses a Queue queue plus a Set seen that records if a node has ever been enqueued. This pushes all the work to the neighbors function - in our Python implementation, all the code after while queue: is template code, except for if node in dead: continue.

As for the neighbors function, for each position in the lock i = 0, 1, 2, 3, for each of the turns d = -1, 1, we determine the value of the lock after the i-th wheel has been turned in the direction d.

Care should be taken in our algorithm, as the graph does not have an edge unless both nodes are not in deadends. If our neighbors function checks only the target for being in deadends, we also need to check whether '0000' is in deadends at the beginning. In our implementation, we check at the visitor level so as to neatly handle this problem in all cases.

In Java, our implementation also inlines the neighbors function for convenience, and uses null inputs in the queue to represent a layer change. When the layer changes, we depth++ our global counter, and queue.peek() != null checks if there are still nodes enqueued.

class Solution(object):
    def openLock(self, deadends, target):
        def neighbors(node):
            for i in xrange(4):
                x = int(node[i])
                for d in (-1, 1):
                    y = (x + d) % 10
                    yield node[:i] + str(y) + node[i+1:]

        dead = set(deadends)
        queue = collections.deque([('0000', 0)])
        seen = {'0000'}
        while queue:
            node, depth = queue.popleft()
            if node == target: return depth
            if node in dead: continue
            for nei in neighbors(node):
                if nei not in seen:
                    seen.add(nei)
                    queue.append((nei, depth+1))
        return -1

Complexity Analysis

Time Complexity: $O(N^2 * \mathcal{A}^N + D)$ where $\mathcal{A}$ is the number of digits in our alphabet, $N$ is the number of digits in the lock, and $D$ is the size of deadends. We might visit every lock combination, plus we need to instantiate our set dead. When we visit every lock combination, we spend $O(N^2)$ time enumerating through and constructing each node.
Space Complexity: $O(\mathcal{A}^N + D)$ , for the queue and the set dead.

 class Solution:
    def openLock(self, deadends: List[str], target: str) -> int:
#         q = deque()
#         visited = set(deadends)

#         if '0000' in visited:
#             return -1


#         q.append(["0000", 0])


#         def combination(lock):
#             res = []
#             for i in range(4):
#                 num = str((int(lock[i])+1) % 10)
#                 comb = lock[:i] + num + lock[i+1:]
#                 res.append(comb)

#                 num = str((int(lock[i])-1 + 10) % 10)
#                 comb = lock[:i] + num + lock[i+1:]
#                 res.append(comb)

#             return res




#         while q:
#             lock, step = q.popleft()

#             if lock == target:
#                 return step

#             for child in combination(lock):
#                 if child not in visited:
#                     visited.add(child)
#                     q.append([child, step+1])


#         return -1
        q = deque()
        visited = set(deadends)

        if '0000' in visited:
            return -1


        q.append(["0000", 0])


        def combination(lock):
            #res = []
            for i in range(4):
                for operation in (1, -1):

                    num = str((int(lock[i])+ operation + 10) % 10)
                    comb = lock[:i] + num + lock[i+1:]
                    yield comb

        while q:
            lock, step = q.popleft()

            if lock == target:
                return step

            for child in combination(lock):
                if child not in visited:
                    visited.add(child)
                    q.append([child, step+1])


        return -1

752. Open The Lock

Approach #1: Breadth-First Search [Accepted]

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