feat: word-break-ii challenge

Author: IbatoLionDev

3-HARD/Python/10-word-break/proposed-solution/__pycache__/word_break.cpython-312.pyc

@@ -0,0 +1,17 @@
+# Challenge: Given a string and a dictionary, return all possible valid sentences that can be formed.
+# This challenge requires recursion with memoization to avoid redundant solutions and explore multiple paths.
+# Use object-oriented programming and follow the DRY principle.
+
+from word_break import WordBreak
+
+def main():
+    s = "catsanddog"
+    word_dict = {"cat", "cats", "and", "sand", "dog"}
+    wb = WordBreak(s, word_dict)
+    sentences = wb.word_break()
+    print("Possible sentences:")
+    for sentence in sentences:
+        print(sentence)
+
+if __name__ == "__main__":
+    main()

3-HARD/Python/10-word-break/proposed-solution/main.py

@@ -0,0 +1,25 @@
+from typing import List, Set, Dict
+
+class WordBreak:
+    def __init__(self, s: str, word_dict: Set[str]):
+        self.s = s
+        self.word_dict = word_dict
+        self.memo: Dict[int, List[str]] = {}
+
+    def word_break(self, start: int = 0) -> List[str]:
+        if start == len(self.s):
+            return [""]
+
+        if start in self.memo:
+            return self.memo[start]
+
+        sentences = []
+        for end in range(start + 1, len(self.s) + 1):
+            word = self.s[start:end]
+            if word in self.word_dict:
+                for subsentence in self.word_break(end):
+                    sentence = word + (" " + subsentence if subsentence else "")
+                    sentences.append(sentence)
+
+        self.memo[start] = sentences
+        return sentences