Open the lock

Time: O(KxN^K + D); Space: O(KxN^K + D); 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.

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

Constraints:

  • 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’.

Hint:

  1. 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.

[4]:

class Solution1(object):
    """
    Time: O(k * n^k + d), n is the number of alphabets,
                          k is the length of target,
                          d is the size of deadends
    Space: O(k * n^k + d)

    """
    def openLock(self, deadends, target):
        """
        :type deadends: List[str]
        :type target: str
        :rtype: int
        """
        dead = set(deadends)
        q = ["0000"]
        lookup = {"0000"}
        depth = 0
        while q:
            next_q = []
            for node in q:
                if node == target:
                    return depth
                if node in dead:
                    continue
                for i in range(4):
                    n = int(node[i])
                    for d in (-1, 1):
                        nn = (n+d) % 10
                        neighbor = node[:i] + str(nn) + node[i+1:]
                        if neighbor not in lookup:
                            lookup.add(neighbor)
                            next_q.append(neighbor)
            q = next_q
            depth += 1
        return -1

[5]:

s = Solution1()
deadends = ["0201","0101","0102","1212","2002"]
target = "0202"
assert s.openLock(deadends, target) == 6
deadends = ["8888"]
target = "0009"
assert s.openLock(deadends, target) == 1
deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"]
target = "8888"
assert s.openLock(deadends, target) == -1
deadends = ["0000"]
target = "8888"
assert s.openLock(deadends, target) == -1