Share my 130 ms Python solution

  • 10

    Main idea:

    1. Use character flipping

    2. Two-end BFS

    3. defaultdict(list) for easy writing to keep track of paths

    I also use set intersection to determine if we are done

    from collections import defaultdict
    class Solution:
        # @param start, a string
        # @param end, a string
        # @param dict, a set of string
        # @return a list of lists of string
        def findLadders(self, start, end, dict):
            wordLen = len(start)
            front, back = defaultdict(list), defaultdict(list)
            # remove start from dict, add end to dict if it is not there
            if end not in dict:
            forward, result = True, []
            while front:
                # get all valid transformations
                nextSet = defaultdict(list)
                for word, paths in front.items():
                    for index in range(wordLen):
                        for ch in 'abcdefghijklmnopqrstuvwxyz':
                            nextWord = word[:index] + ch + word[index+1:]
                            if nextWord in dict:
                                # update paths
                                if forward:
                                    # append next word to path
                                    nextSet[nextWord].extend([path + [nextWord] for path in paths])
                                    # add next word in front of path
                                    nextSet[nextWord].extend([[nextWord] + path for path in paths])
                front = nextSet
                common = set(front) & set(back)
                if common:
                    # path is through
                    if not forward:
                        # switch front and back if we were searching backward
                        front, back = back, front
                    result.extend([head + tail[1:] for word in common for head in front[word] for tail in back[word]])
                    return result
                if len(front) > len(back):
                    # swap front and back for better performance (smaller nextSet)
                    front, back, forward = back, front, not forward
                # remove transformations from wordDict to avoid cycles
                dict -= set(front)
            return []

  • 0

    what if start == end? Does it still work?

  • 0

    It will create duplicates. If there is such test cases, we need to add special treatment

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.