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: 6 Explanation: 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: 1 Explanation: 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: -1 Explanation: We can't reach the target without getting stuck.
Example 4:
Input: deadends = ["0000"], target = "8888" Output: -1
Note:
deadends
will be in the range [1, 500]
.target
will not be in the list deadends
.deadends
and the string target
will be a string of 4 digits from the 10,000 possibilities '0000'
to '9999'
.Intuition
\nWe 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
\nTo 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
\nTime Complexity: where is the number of digits in our alphabet, is the number of digits in the lock, and is the size of deadends
. We might visit every lock combination, plus we need to instantiate our set dead
.
Space Complexity: , for the queue
and the set dead
.
Analysis written by: @awice.
\n