Added Day-30 Solution

Author: Gangadharbhuvan

Day-30_Word_Break_II.py

@@ -0,0 +1,60 @@
+'''
+    Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.
+
+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 = "catsanddog"
+wordDict = ["cat", "cats", "and", "sand", "dog"]
+Output:
+[
+  "cats and dog",
+  "cat sand dog"
+]
+Example 2:
+
+Input:
+s = "pineapplepenapple"
+wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
+Output:
+[
+  "pine apple pen apple",
+  "pineapple pen apple",
+  "pine applepen apple"
+]
+Explanation: Note that you are allowed to reuse a dictionary word.
+Example 3:
+
+Input:
+s = "catsandog"
+wordDict = ["cats", "dog", "sand", "and", "cat"]
+Output:
+[]
+
+'''
+
+class Solution:
+    def wordBreak(self, s, wordDict):
+        wordSet = set(wordDict)
+        n = len(s)
+        dp_solution = [[] for _ in range(n)] + [[""]]
+        dp = [0] * n + [1]
+        
+        for k in range(n):
+            for j in range(k,-1,-1):
+                if s[j: k + 1] in wordSet:
+                    dp[k] = max(dp[k], dp[j-1])
+
+        if dp[-2] == 0: return []
+
+        for k in range(n):
+            for j in range(k,-1,-1):
+                if s[j: k + 1] in wordSet:
+                    for sol in dp_solution[j-1]:
+                        dp_solution[k].append(sol + " " + s[j: k + 1])
+                        
+        return [s[1:] for s in dp_solution[-2]]