solve: designAddAndSearchWordsDataStructure

Author: yolophg

design-add-and-search-words-data-structure/yolophg.py

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+# Time Complexity:  O(n) - where n is the length of the word(addWord)
+# Space Complexity: O(N) - where N is the total number of characters inserted
+
+class WordDictionary:
+    def __init__(self):
+        # using a trie (prefix tree) to store all the words
+        self.trie = dict()
+
+    def addWord(self, word: str) -> None:
+        # start from the root of the trie
+        trie = self.trie
+        for letter in word:
+            # if this letter isn't already in the current trie level, add it
+            if letter not in trie:
+                trie[letter] = dict()
+            # move one level deeper
+            trie = trie[letter]
+        # mark the end of the word with a special null character
+        trie['\0'] = dict()
+
+    def internal_search(self, trie: dict, index: int, word: str) -> bool:
+        if index == len(word):
+            # check if this path ends a valid word
+            return '\0' in trie  
+
+        letter = word[index]
+        
+        # if hit a '.', gotta try all possible paths from here
+        if letter == '.':
+            for child in trie.values():
+                if self.internal_search(child, index + 1, word):
+                    return True
+            return False
+        else:
+            # if the letter exists in the current trie level, keep going
+            if letter in trie:
+                return self.internal_search(trie[letter], index + 1, word)
+            else:
+                return False
+
+    def search(self, word: str) -> bool:
+        # start the recursive search from index 0 and the root
+        return self.internal_search(self.trie, 0, word)