# Time:  O((25n)^n)
# Space: O((25n)^n)
#
# Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
# 
# Only one letter can be changed at a time
# Each intermediate word must exist in the dictionary
# For example,
# 
# Given:
# start = "hit"
# end = "cog"
# dict = ["hot","dot","dog","lot","log"]
# As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
# return its length 5.
# 
# Note:
# Return 0 if there is no such transformation sequence.
# All words have the same length.
# All words contain only lowercase alphabetic characters.
#

# BFS
class Solution:
    # @param start, a string
    # @param end, a string
    # @param dict, a set of string
    # @return an integer
    def ladderLength(self, start, end, dict):
        distance, cur, visited = 0, [start], set([start])
        dict.add(end)
        
        while cur:
            next = []
            for word in cur:
                if word == end:
                    return distance + 1
                for i in xrange(len(word)):
                    for j in 'abcdefghijklmnopqrstuvwxyz':
                        candidate = word[:i] + j + word[i + 1:]
                        if candidate not in visited and candidate in dict:
                            next.append(candidate)
                            visited.add(candidate)
            distance += 1
            cur = next
        
        return 0
    
if __name__ == "__main__":
    print Solution().ladderLength("hit", "cog", set(["hot","dot","dog","lot","log"]))

# Time:  O(nd), n is length of string, d is size of dictionary
# Space: O(d)
#
# Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
# 
# Only one letter can be changed at a time
# Each intermediate word must exist in the dictionary
# For example,
# 
# Given:
# start = "hit"
# end = "cog"
# dict = ["hot","dot","dog","lot","log"]
# As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
# return its length 5.
# 
# Note:
# Return 0 if there is no such transformation sequence.
# All words have the same length.
# All words contain only lowercase alphabetic characters.
#

# BFS
class Solution:
    # @param start, a string
    # @param end, a string
    # @param dict, a set of string
    # @return an integer
    def ladderLength(self, start, end, dict):
        distance, cur, visited = 0, [start], set([start])
        dict.add(end)
        
        while cur:
            next = []
            for word in cur:
                if word == end:
                    return distance + 1
                for i in xrange(len(word)):
                    for j in 'abcdefghijklmnopqrstuvwxyz':
                        candidate = word[:i] + j + word[i + 1:]
                        if candidate not in visited and candidate in dict:
                            next.append(candidate)
                            visited.add(candidate)
            distance += 1
            cur = next
        
        return 0
    
if __name__ == "__main__":
    print Solution().ladderLength("hit", "cog", set(["hot","dot","dog","lot","log"]))

Update word-ladder.py