<p>Given an <code>m x n</code> <code>board</code>&nbsp;of characters and a list of strings <code>words</code>, return <em>all words on the board</em>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search1.jpg" style="width: 322px; height: 322px;" />
<pre>
<strong>Input:</strong> board = [[&quot;o&quot;,&quot;a&quot;,&quot;a&quot;,&quot;n&quot;],[&quot;e&quot;,&quot;t&quot;,&quot;a&quot;,&quot;e&quot;],[&quot;i&quot;,&quot;h&quot;,&quot;k&quot;,&quot;r&quot;],[&quot;i&quot;,&quot;f&quot;,&quot;l&quot;,&quot;v&quot;]], words = [&quot;oath&quot;,&quot;pea&quot;,&quot;eat&quot;,&quot;rain&quot;]
<strong>Output:</strong> [&quot;eat&quot;,&quot;oath&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> board = [[&quot;a&quot;,&quot;b&quot;],[&quot;c&quot;,&quot;d&quot;]], words = [&quot;abcb&quot;]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 12</code></li>
	<li><code>board[i][j]</code> is a lowercase English letter.</li>
	<li><code>1 &lt;= words.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>words</code> are unique.</li>
</ul>

class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_word = False
        self.ref = 0

    
class Trie:
    def __init__(self):
        self.root = TrieNode()
    
    def add_word(self, word):
        cur = self.root
        for ch in word:
            if ch not in cur.children:
                new_node = TrieNode()
                cur.children[ch] = new_node
            cur = cur.children[ch]
            cur.ref += 1
        cur.is_word = True
    
    def remove_word(self, word):
        cur = self.root
        for ch in word:
            cur = cur.children[ch]
            cur.ref -= 1
        cur.is_word = False

        
class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        if not board or not board[0]:
            return []
        ans = []
        trie = Trie()
        for word in words:
            trie.add_word(word)
        
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                if board[i][j] in trie.root.children:
                    # do not write the visited set wrong
                    visited = set([(i, j)])
                    self.helper(trie, board, i, j, ans, trie.root.children[board[i][j]], visited, m, n, board[i][j])
        return ans

    def helper(self, trie, board, x, y, ans, cur_trie, visited, m, n, word):
        if cur_trie.is_word:
            ans.append(word)
            trie.remove_word(word)
            # !!do not need to return here!!
            #return
        
        for dx, dy in [(0, 1), (0, -1), (-1, 0), (1, 0)]:
            new_x, new_y = x + dx, y + dy
            if 0 <= new_x < m and 0 <= new_y < n and (new_x, new_y) not in visited:
                # in trie_node
                letter = board[new_x][new_y]
                if letter in cur_trie.children and cur_trie.children[letter].ref:
                    word += letter
                    visited.add((new_x, new_y))
                    self.helper(trie, board, new_x, new_y, ans, cur_trie.children[letter], visited, m, n, word)
                    word = word[:-1]
                    visited.remove((new_x, new_y))


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