<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):
        # Initialize a TrieNode with an empty dictionary for children nodes,
        # a boolean flag to indicate if it represents the end of a word,
        # and a reference counter.
        self.children = {}
        self.isWord = False
        self.refs = 0

    def addWord(self, word):
        # Method to add a word to the trie.
        # Traverse the trie, creating nodes as necessary.
        # Increment the reference counter for each visited node.
        cur = self
        cur.refs += 1
        for c in word:
            if c not in cur.children:
                cur.children[c] = TrieNode()
            cur = cur.children[c]
            cur.refs += 1
        cur.isWord = True

    def removeWord(self, word):
        # Method to remove a word from the trie.
        # Decrement the reference counter for each visited node.
        cur = self
        cur.refs -= 1
        for c in word:
            if c in cur.children:
                cur = cur.children[c]
                cur.refs -= 1


class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        # Trie implementation to efficiently search words on the board.
        root = TrieNode()
        # Add all words from the dictionary to the trie.
        for w in words:
            root.addWord(w)

        rows, cols = len(board), len(board[0])
        result, visit = set(), set()

        def dfs(r, c, node, word):
            # Depth-first search to explore possible word paths on the board.
            # Check boundary conditions, character existence, and node references.
            if (
                r not in range(rows) 
                or c not in range(cols)
                or board[r][c] not in node.children
                or node.children[board[r][c]].refs < 1
                or (r, c) in visit
            ):
                return

            visit.add((r, c))
            node = node.children[board[r][c]]
            word += board[r][c]
            if node.isWord:
                # If a word is found, add it to the result set,
                # mark it as not a word in the trie, and remove it.
                node.isWord = False
                result.add(word)
                root.removeWord(word)

            # Explore all four directions recursively.
            dfs(r + 1, c, node, word)
            dfs(r - 1, c, node, word)
            dfs(r, c + 1, node, word)
            dfs(r, c - 1, node, word)
            visit.remove((r, c))

        # Iterate through each cell on the board to start DFS.
        for r in range(rows):
            for c in range(cols):
                dfs(r, c, root, "")

        return list(result)
20000