O(n^2) time and O(n) space using Queue.

Author: 149ps

139. Word Break/139. Word Break.py

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+"""
+Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
+
+Note:
+
+The same word in the dictionary may be reused multiple times in the segmentation.
+You may assume the dictionary does not contain duplicate words.
+Example 1:
+
+Input: s = "leetcode", wordDict = ["leet", "code"]
+Output: true
+Explanation: Return true because "leetcode" can be segmented as "leet code".
+Example 2:
+
+Input: s = "applepenapple", wordDict = ["apple", "pen"]
+Output: true
+Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
+             Note that you are allowed to reuse a dictionary word.
+Example 3:
+
+Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
+Output: false
+"""
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        visited = set()
+        q = collections.deque([0])
+        while q:
+            start = q.popleft()
+            if start not in visited:
+                for end in range(start+1,len(s)+1):
+                    if s[start:end] in wordDict:
+                        q.append(end)
+                        if end == len(s):
+                            return True
+                visited.add(start)
+        return False