<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.end = False
    
    def add(self, word):
        curr = self
        for i in word:
            if i not in curr.children:
                curr.children[i] = TrieNode()
            curr = curr.children[i]
        curr.end = True
    
    def prune(self, word):
        # Start pruning from the root node
        def _prune(node, index):
            # If we've reached the end of the word
            if index == len(word):
                # Mark the end of the word as False
                node.end = False
                # If node has no children, return True to prune this node
                return len(node.children) == 0
            
            # Get the current character
            char = word[index]
            # Recursively prune the children
            if char in node.children and _prune(node.children[char], index + 1):
                # If child can be pruned, delete it
                del node.children[char]
                # Return True if the current node can also be pruned
                return not node.end and len(node.children) == 0
            
            return False
        
        _prune(self, 0)

class Solution:
    def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
        trie = TrieNode()
        for word in words:
            trie.add(word)
        
        ret = set()
        row, col = len(board), len(board[0])
        visit = set()
        
        def backtrack(i, j, node, curr):
            if i not in range(row) or j not in range(col) or board[i][j] not in node.children or (i, j) in visit:
                return  
            
            curr += board[i][j]
            nextNode = node.children[board[i][j]]

            if nextNode.end:
                ret.add(curr)
                trie.prune(curr)  # Prune the word from the Trie
            
            visit.add((i, j))

            # Explore the neighbors
            backtrack(i - 1, j, nextNode, curr)
            backtrack(i + 1, j, nextNode, curr)
            backtrack(i, j + 1, nextNode, curr)
            backtrack(i, j - 1, nextNode, curr)

            visit.remove((i, j))
        
        # Start backtracking from each cell in the grid
        for i in range(row):
            for j in range(col):
                backtrack(i, j, trie, "")

        return list(ret)

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