<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 Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        
        m, n = len(board), len(board[0])

        def ins(w, d, idx):
            
            if not w:
                if '#' not in d:
                    d['#'] = set()
                
                d['#'].add(idx)

                return d
            
            if w[0] not in d:
                d[w[0]] = {}

            d[w[0]] = ins(w[1:], d[w[0]], idx)        
            return d
        
        d = {}

        for idx, word in enumerate(words):
            d = ins(word, d, idx)
        
        res = set()
        deleted = set()

        def delete(idx):
            nonlocal d

            def _delete(w, d, idx):
                
                if not w:
                    d['#'].discard(idx)
                    if not d['#']:
                        del d['#']
                    return d
                
                d[w[0]] = _delete(w[1:], d[w[0]], idx)

                if not d[w[0]]:
                    del d[w[0]]
                
                return d
            

            d = _delete(words[idx], d, idx)
            if not d:
                d = {}
            
                


        
        def dfs(i, j, d):
            
            if '#' in d:
                nonlocal res, deleted
                res = res.union(d['#'])

                for del_idx in d['#'].copy():
                    if del_idx not in deleted:
                        delete(del_idx)
                        deleted.add(del_idx)
            
            for di, dj in [(0, 1), (1, 0), (-1, 0), (0, -1)]:
                x, y = di + i, dj + j

                if not (0 <= x < m and 0 <= y < n) or (x, y) in vis or board[x][y] not in d:
                    continue
                
                vis.add((x, y))
                dfs(x, y, d[board[x][y]])
                vis.discard((x, y))
            
        vis = set()

        for i in range(m):
            for j in range(n):

                if board[i][j] in d:
                    vis.add((i, j))
                    dfs(i, j, d[board[i][j]])
                    vis.discard((i, j))
        
        return [words[i] for i in res]





            
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