From 32cb62a004e3c92c79adb8fb3ac096b460f4075b Mon Sep 17 00:00:00 2001
From: Nishant Nirmal <nirmalnishant645@gmail.com>
Date: Thu, 9 Apr 2020 21:17:45 +0530
Subject: [PATCH] Word Break

---
 0139-Word-Break.py | 43 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 0139-Word-Break.py

diff --git a/0139-Word-Break.py b/0139-Word-Break.py
new file mode 100644
index 0000000..c0caa79
--- /dev/null
+++ b/0139-Word-Break.py
@@ -0,0 +1,43 @@
+'''
+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
+'''
+
+# Method 1
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        ok = [True]
+        for i in range(1, len(s) + 1):
+            ok += any(ok[j] and s[j : i] in wordDict for j in range(i)),
+        return ok[-1]
+
+# Method 2
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        dp = [False] * (len(s) + 1)
+        dp[0] = True
+        for i in range(len(s)):
+            if dp[i]:
+                for j in range(i + 1, len(s) + 1):
+                    if s[i : j] in wordDict:
+                        dp[j] = True     
+        return dp[-1]