Word Break

Author: anantkaushik

139-word-break.py

@@ -0,0 +1,42 @@
+"""
+Problem Link: https://leetcode.com/problems/word-break/
+
+Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a 
+space-separated sequence of one or more dictionary words.
+Note that the same word in the dictionary may be reused multiple times in the segmentation.
+
+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
+ 
+Constraints:
+1 <= s.length <= 300
+1 <= wordDict.length <= 1000
+1 <= wordDict[i].length <= 20
+s and wordDict[i] consist of only lowercase English letters.
+All the strings of wordDict are unique.
+"""
+# TLE - O (2^N)
+class Solution:
+ def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+ return self.helper(s, set(wordDict))
+ 
+ def helper(self, s, words):
+ if not s:
+ return True
+ 
+ for index in range(1, len(s)+1):
+ if s[:index] in words and self.helper(s[index:], words):
+ return True
+ return False