Word Ladder/ Word Break II

Author: gavinfish

127 Word Ladder.py

@@ -0,0 +1,51 @@
+'''
+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:
+
+Only one letter can be changed at a time
+Each intermediate word must exist in the word list
+For example,
+
+Given:
+beginWord = "hit"
+endWord = "cog"
+wordList = ["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.
+'''
+
+class Solution(object):
+    def ladderLength(self, beginWord, endWord, wordList):
+        """
+        :type beginWord: str
+        :type endWord: str
+        :type wordList: Set[str]
+        :rtype: int
+        """
+        wordList.add(endWord)
+        cur_level = [beginWord]
+        next_level = []
+        depth = 1
+        n = len(beginWord)
+        while cur_level:
+            for item in cur_level:
+                if item == endWord:
+                    return depth
+                for i in range(n):
+                    for c in 'abcdefghijklmnopqrstuvwxyz':
+                        word = item[:i] + c + item[i + 1:]
+                        if word in wordList:
+                            wordList.remove(word)
+                            next_level.append(word)
+            depth += 1
+            cur_level = next_level
+            next_level = []
+        return 0
+
+
+if __name__ == "__main__":
+    assert Solution().ladderLength("hit", "cog", {"hot", "dot", "dog", "lot", "log"}) == 5

140 Word Break II.py

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+'''
+Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
+
+Return all such possible sentences.
+
+For example, given
+s = "catsanddog",
+dict = ["cat", "cats", "and", "sand", "dog"].
+
+A solution is ["cats and dog", "cat sand dog"].
+'''
+
+import collections
+
+
+class Solution(object):
+    def wordBreak(self, s, wordDict):
+        """
+        :type s: str
+        :type wordDict: Set[str]
+        :rtype: List[str]
+        """
+        dic = collections.defaultdict(list)
+
+        def dfs(s):
+            if not s:
+                return [None]
+            if s in dic:
+                return dic[s]
+            res = []
+            for word in wordDict:
+                n = len(word)
+                if s[:n] == word:
+                    for r in dfs(s[n:]):
+                        if r:
+                            res.append(word + " " + r)
+                        else:
+                            res.append(word)
+            dic[s] = res
+            return res
+
+        return dfs(s)
+
+
+if __name__ == "__main__":
+    assert Solution().wordBreak("catsanddog", {"cat", "cats", "and", "sand", "dog"}) == ['cat sand dog', 'cats and dog']