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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.

Given a `target`

representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.

**Example 1:**

Input:deadends = ["0201","0101","0102","1212","2002"], target = "0202"Output:6Explanation:A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202". Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid, because the wheels of the lock become stuck after the display becomes the dead end "0102".

**Example 2:**

Input:deadends = ["8888"], target = "0009"Output:1Explanation:We can turn the last wheel in reverse to move from "0000" -> "0009".

**Example 3:**

Input:deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"Output:-1Explanation:We can't reach the target without getting stuck.

**Example 4:**

Input:deadends = ["0000"], target = "8888"Output:-1

**Note:**

- The length of
`deadends`

will be in the range`[1, 500]`

. `target`

will not be in the list`deadends`

.- Every string in
`deadends`

and the string`target`

will be a string of 4 digits from the 10,000 possibilities`'0000'`

to`'9999'`

.

b'

\n\n#### Approach #1: Breadth-First Search [Accepted]

\n

\n

'
**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.

**Complexity Analysis**

- \n
- \n
Time Complexity: where is the number of digits in our alphabet, is the number of digits in the lock, and is the size of

\n`deadends`

. We might visit every lock combination, plus we need to instantiate our set`dead`

. \n - \n
Space Complexity: , for the

\n`queue`

and the set`dead`

. \n

\n

Analysis written by: @awice.

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