Add back word break II original solution

Author: SocialfiPanda

Word Break II.py

@@ -24,3 +24,34 @@ def backtrack(self, s, trace, start, path, result):
         for i in xrange(start, len(s)):
             if trace[start][i]:
                 self.backtrack(s, trace, i + 1, path + [s[start:i+1]], result)
+
+class Solution:  
+    """ The alternative solution is to precheck before actually break the words.
+    """
+    def canBreak(self, s, dict):
+        possible = []
+        for i in range(len(s)):
+            if s[:i + 1] in dict:
+                possible.append(True)
+            else:
+                found = False
+                for j in range(i):
+                    if possible[j] == True and s[j + 1: i + 1] in dict:
+                        found = True
+                        break
+                possible.append(found)
+        return possible[len(s) - 1]
+        
+    def wordBreak(self, s, dict):
+        result = {}
+        if not self.canBreak(s, dict):
+            return []
+        for i in range(len(s)):
+            result[s[:i + 1]] = []
+            if s[:i + 1] in dict:
+                result[s[:i + 1]] = [s[:i + 1]]
+            for j in range(i):
+                if s[:j + 1] in result and s[j + 1: i + 1] in dict:
+                    for k in result[s[:j + 1]]:
+                        result[s[:i + 1]].append(k + " " + s[j + 1: i + 1])
+        return result[s]