<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 trie_node:
    def __init__(self,):
        self.children = {}
        self.end_of_word = False
        self.ref = 0
    
    def insert(self, word):
        cur = self
        cur.ref += 1
        for char in word:
            if char not in cur.children:
                cur.children[char] = trie_node()
            cur = cur.children[char]
            cur.ref += 1
        cur.end_of_word = True

    def delete(self, word):
        cur = self
        cur.ref -= 1
        for char in word:
            cur = cur.children[char]
            cur.ref -= 1
        cur.end_of_word = False

class Solution:
    def __init__(self,):
        self.trie = trie_node()
    
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        rows, cols = len(board), len(board[0])

        for word in words:
            self.trie.insert(word)

        path = set()
        result = set()
        def backtrack(row, col, node, prefix):
            if( row<0 or col<0 or
                row>=rows or col>=cols or
                node.ref<=0 or 
                board[row][col] not in node.children or
                (row,col) in path) :
                return 
            
            path.add((row,col))
            prefix+=board[row][col]
            node = node.children[board[row][col]]

            if node.end_of_word:
                result.add(prefix)
                self.trie.delete(prefix)
            
            temp = (backtrack(row-1, col, node, prefix),
                    backtrack(row, col-1, node, prefix),
                    backtrack(row+1, col, node, prefix),
                    backtrack(row, col+1, node, prefix))
            path.remove((row,col))

            return temp
        
        for row in range(rows):
            for col in range(cols):
                backtrack(row, col, self.trie, "")
        print(result)
        return list(result)
19400