solve: wordSearchIi

Author: yolophg

word-search-ii/yolophg.py

@@ -0,0 +1,45 @@
+# Time Complexity: O(m * n * 4^l) -> each cell can explore up to 4 directions recursively, where l is the max word length.
+# Space Complexity: O(w * l) -> storing words in a dictionary-based Trie.
+
+class Solution:
+    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:    
+        # trie-like dictionary to store words    
+        d = {} 
+        
+        # build the trie
+        for word in words:
+            cur = d
+            for c in word:
+                if c not in cur:
+                     # create a new node
+                    cur[c] = {} 
+                cur = cur[c]
+            # mark the end of the word
+            cur["*"] = word  
+        
+        # right, down, up, left
+        directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]  
+        
+        # backtracking function
+        def dfs(i, j, cur, seen):
+            result = set()
+            if "*" in cur:
+                # found a word, add it
+                result = {cur["*"]}  
+            
+            for x, y in directions:
+                ni, nj = i + x, j + y
+                if 0 <= ni < len(board) and 0 <= nj < len(board[0]) and (ni, nj) not in seen and board[ni][nj] in cur:
+                    result.update(dfs(ni, nj, cur[board[ni][nj]], seen | {(ni, nj)}))
+            
+            return result
+        
+        result = set()
+        
+        # start dfs search from every cell in the board
+        for i in range(len(board)):
+            for j in range(len(board[0])):
+                if board[i][j] in d:
+                    result.update(dfs(i, j, d[board[i][j]], {(i, j)}))
+        
+        return list(result)