Word break - Leetcode problem.

Author: sanjaypradeep

DynamicProgramming/wordBreak.py

@@ -0,0 +1,107 @@
+"""
+Task:
+Given a string and a list of words, return true if the string can be
+segmented into a space-separated sequence of one or more words.
+
+Note that the same word may be reused
+multiple times in the segmentation.
+
+Implementation notes: Trie + Dynamic programming up -> down.
+The Trie will be used to store the words. It will be useful for scanning
+available words for the current position in the string.
+
+Leetcode:
+https://leetcode.com/problems/word-break/description/
+
+Runtime: O(n * n)
+Space: O(n)
+"""
+
+import functools
+from typing import Any
+
+
+def word_break(string: str, words: list[str]) -> bool:
+    """
+    Return True if numbers have opposite signs False otherwise.
+
+    >>> word_break("applepenapple", ["apple","pen"])
+    True
+    >>> word_break("catsandog", ["cats","dog","sand","and","cat"])
+    False
+    >>> word_break("cars", ["car","ca","rs"])
+    True
+    >>> word_break('abc', [])
+    False
+    >>> word_break(123, ['a'])
+    Traceback (most recent call last):
+        ...
+    ValueError: the string should be not empty string
+    >>> word_break('', ['a'])
+    Traceback (most recent call last):
+        ...
+    ValueError: the string should be not empty string
+    >>> word_break('abc', [123])
+    Traceback (most recent call last):
+        ...
+    ValueError: the words should be a list of non-empty strings
+    >>> word_break('abc', [''])
+    Traceback (most recent call last):
+        ...
+    ValueError: the words should be a list of non-empty strings
+    """
+
+    # Validation
+    if not isinstance(string, str) or len(string) == 0:
+        raise ValueError("the string should be not empty string")
+
+    if not isinstance(words, list) or not all(
+        isinstance(item, str) and len(item) > 0 for item in words
+    ):
+        raise ValueError("the words should be a list of non-empty strings")
+
+    
+    # Build trie
+    trie: dict[str, Any] = {}
+    word_keeper_key = "WORD_KEEPER"
+
+    for word in words:
+        trie_node = trie
+        for c in word:
+            if c not in trie_node:
+                trie_node[c] = {}
+
+            trie_node = trie_node[c]
+
+        trie_node[word_keeper_key] = True
+
+    len_string = len(string)
+
+    # Dynamic programming method
+    @functools.cache
+    def is_breakable(index: int) -> bool:
+        """
+        >>> string = 'a'
+        >>> is_breakable(1)
+        True
+        """
+        if index == len_string:
+            return True
+
+        trie_node = trie
+        for i in range(index, len_string):
+            trie_node = trie_node.get(string[i], None)
+
+            if trie_node is None:
+                return False
+
+            if trie_node.get(word_keeper_key, False) and is_breakable(i + 1):
+                return True
+
+        return False
+
+    return is_breakable(0)
+
+
+if __name__ == "__main__":
+    print(word_break("applepenapple", ["apple", "pen"]))