# 212 Word Search II

Given an `m x n` board of characters and a list of strings words, return all words on the board.

Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example 1:

``````Input: board = [
["o","a","a","n"],
["e","t","a","e"],
["i","h","k","r"],
["i","f","l","v"]
], words = ["oath","pea","eat","rain"]
Output: ["eat","oath"]
``````

Example 2:

``````Input: board = [["a","b"],["c","d"]], words = ["abcb"]
Output: []
``````

Constraints:

• `m == board.length`
• `n == board[i].length`
• `1 <= m, n <= 12`
• `board[i][j]` is a lowercase English letter.
• `1 <= words.length <= 3 * 104`
• `1 <= words[i].length <= 10`
• `words[i]` consists of lowercase English letters.
• All the strings of `words` are unique.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 `````` ``````class Solution: def findWords( self, board: List[List[str]], words: List[str] ) -> List[str]: def prune_shitty_trie(root, word): prune_candidate = (root, word[0]) current = root i = 0 while i < len(word): char = word[i] if current[char]["is_word"] or len(current[char]) > 2: # can't prune anything above and including current[char] if i + 1 >= len(word): return prune_candidate = (current[char], word[i + 1]) current = current[char] i += 1 if len(current) == 1: del prune_candidate[0][prune_candidate[1]] def dfs(r, c, node, accumulated): if r < 0 or r >= m or c < 0 or c >= n: return if board[r][c] not in node: return char = board[r][c] board[r][c] = "#" accumulated += char node = node[char] if node["is_word"]: res.append(accumulated) node["is_word"] = False prune_shitty_trie(root, accumulated) # 4 directions for (dr, dc) in [(1, 0), (-1, 0), (0, 1), (0, -1)]: dfs(r + dr, c + dc, node, accumulated) board[r][c] = char m, n = len(board), len(board[0]) # Build trie root = {"is_word": False} for word in words: current = root for char in word: current = current.setdefault(char, {"is_word": False}) current["is_word"] = True res = [] for r in range(m): for c in range(n): dfs(r, c, root, '') return res``````