add word_break dynamic approach up -> down.

dynamic_programming/word_break.py

@@ -0,0 +1,105 @@
+"""
+Author  : Alexander Pantyukhin
+Date    : December 12, 2022
+
+Task:
+Given a string s and a dictionary of strings wordDict,
+return true if s can be segmented into a space-separated
+sequence of one or more dictionary words.
+
+Note that the same word in the dictionary may be reused
+multiple times in the segmentation.
+
+Implementation notes: Trie + Dynamic programming up -> down.
+The Trie keeps all wordDict 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)
+"""
+
+from functools import lru_cache
+
+def wordBreak(s: str, word_dict: list[str]) -> bool:
+    """
+    Return True if numbers have opposite signs False otherwise.
+
+    >>> wordBreak("applepenapple", ["apple","pen"])
+    True
+    >>> wordBreak("catsandog", ["cats","dog","sand","and","cat"])
+    False
+    >>> wordBreak("cars", ["car","ca","rs"])
+    True
+    >>> wordBreak("aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"])
+    False
+    >>> wordBreak('abc', [])
+    False
+    >>> wordBreak(123, ['a'])
+    Traceback (most recent call last):
+        ...
+    ValueError: the s should be not empty string
+    >>> wordBreak('', ['a'])
+    Traceback (most recent call last):
+        ...
+    ValueError: the s should be not empty string
+    >>> wordBreak('abc', [123])
+    Traceback (most recent call last):
+        ...
+    ValueError: the word_dict should a list of non empty string
+    >>> wordBreak('abc', [''])
+    Traceback (most recent call last):
+        ...
+    ValueError: the word_dict should a list of non empty string
+    """
+
+    # Validation
+    if not isinstance(s, str) or len(s) == 0:
+        raise ValueError('the s should be not empty string')
+
+    if not isinstance(word_dict, list) or not all(
+        [isinstance(item, str) and len(item) > 0 for item in word_dict]):
+        raise ValueError('the word_dict should a list of non empty string')
+
+    # Build trie
+    trie = {}
+    WORD_KEEPER = 'WORD_KEEPER'
+    for word in word_dict:
+        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] = True
+
+    len_s = len(s)
+
+    # Dynamic programming method
+    @lru_cache(maxsize=None)
+    def is_breakable(index: int) -> bool:
+        if index == len_s:
+            return True
+
+        trie_node = trie
+        for i in range(index, len_s):
+            trie_node = trie_node.get(s[i], None)
+
+            if trie_node is None:
+                return False
+
+            if trie_node.get(WORD_KEEPER, False) and is_breakable(i + 1):
+                return True
+
+        return False
+
+    return is_breakable(0)
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()