118 lines
3.1 KiB
Markdown
118 lines
3.1 KiB
Markdown
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---
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layout: page
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title: Word Ladder
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last_solved: 2020-07-23
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category: bfs
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leetcode_url: https://leetcode.com/problems/word-ladder
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status: Attempted
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---
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Problem
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-------
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```
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Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:
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Only one letter can be changed at a time.
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Each transformed word must exist in the word list.
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Note:
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Return 0 if there is no such transformation sequence.
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All words have the same length.
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All words contain only lowercase alphabetic characters.
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You may assume no duplicates in the word list.
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You may assume beginWord and endWord are non-empty and are not the same.
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Example 1:
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Input:
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beginWord = "hit",
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endWord = "cog",
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wordList = ["hot","dot","dog","lot","log","cog"]
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Output: 5
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Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
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return its length 5.
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Example 2:
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Input:
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beginWord = "hit"
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endWord = "cog"
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wordList = ["hot","dot","dog","lot","log"]
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Output: 0
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Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.
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```
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Solution
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----------
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I did not solve this myself yet. Just got it from solutions tab and tried to analyse it.
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{% highlight python %}
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def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int:
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# {'*og': ['dog', 'log', 'cog'],
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# '*ot': ['hot', 'dot', 'lot'],
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# 'c*g': ['cog'],
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# 'co*': ['cog'],
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# 'd*g': ['dog'],
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# 'd*t': ['dot'],
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# 'do*': ['dot', 'dog'],
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# 'h*t': ['hot'],
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# 'ho*': ['hot'],
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# 'l*g': ['log'],
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# 'l*t': ['lot'],
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# 'lo*': ['lot', 'log']})
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if endWord not in wordList or not endWord or not beginWord or not wordList:
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return 0
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L = len(beginWord)
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# make adjacency list
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all_combo_dict = defaultdict(list)
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for word in wordList:
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for i in range(L):
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# key is the generic word
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# value is a list of words which have the same intermediate generic word
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all_combo_dict[word[:i] + "*" + word[i+1:]].append(word)
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#pp.pprint(all_combo_dict)
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# Queue for BFS
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queue = collections.deque([(beginWord, 1)]) # (word, level)
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# Visited to make sure we don't repeat processing same word.
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visited = {beginWord: True}
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while queue:
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current_word, level = queue.popleft()
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for i in range(L):
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# intermediate words for current word
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intermediate_word = current_word[:i] + "*" + current_word[i+1:]
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# next states are all the words which share the same intermediate state
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for word in all_combo_dict[intermediate_word]:
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if word==endWord:
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return level+1
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if word not in visited:
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visited[word] = True
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queue.append((word, level+1) )
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all_combo_dict[intermediate_word] = []
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return 0
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{% endhighlight %}
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![image1]()
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