<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]:
        WORD_MARK = '#'
        word_trie = {}
        row = len(board)
        col = len(board[0])
        matched = []
        
        def insert_into_trie(trie, words):
            for word in words:
                node = trie
                for ch in word:
                    node.setdefault(ch, {})
                    node = node[ch]
                node[WORD_MARK] = word

        def search(pos, trie):
            x, y = pos
            char = board[x][y]
            node = trie[char] # Not a miss if a check is made before calling the function

            matched_word = node.pop(WORD_MARK, False) # Popping removes the possibility of adding dupes
            if matched_word: matched.append(matched_word)

            board[x][y] = '$'

            for dx, dy in [(-1, 0), (0, 1), (1, 0), (0, -1)]:
                x_new, y_new = x + dx, y + dy
                if x_new < 0 or x_new >= row or y_new < 0 or y_new >= col:
                    continue
                if board[x_new][y_new] not in node:
                    continue
                search((x_new, y_new), node)

            board[x][y] = char
        
        insert_into_trie(word_trie, words)

        for r in range(row):
            for c in range(col):
                if board[r][c] in word_trie:
                    search((r, c), word_trie)

        return matched
        


17200