854. K-Similar Strings

# Time:  O(n * n!/(c_a!*...*c_z!), n is the length of A, B,
#                                  c_a...c_z is the count of each alphabet,
#                                  n = sum(c_a...c_z)
# Space: O(n * n!/(c_a!*...*c_z!)

import collections


class Solution(object):
    def kSimilarity(self, A, B):
        """
        :type A: str
        :type B: str
        :rtype: int
        """
        def neighbors(s, B):
            for i, c in enumerate(s):
                if c != B[i]:
                    break
            t = list(s)
            for j in xrange(i+1, len(s)):
                if t[j] == B[i]:
                    t[i], t[j] = t[j], t[i]
                    yield "".join(t)
                    t[j], t[i] = t[i], t[j]

        q = collections.deque([A])
        lookup = set()
        result = 0
        while q:
            for _ in xrange(len(q)):
                s = q.popleft()
                if s == B:
                    return result
                for t in neighbors(s, B):
                    if t not in lookup:
                        lookup.add(t)
                        q.append(t)
            result += 1

Solution from kamyu104/LeetCode-Solutions · MIT