Word Search II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0212.Word%20Search%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0212.Word%20Search%20II/images/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>
Trie; Array; String; Backtracking; Matrix
class Trie {
public:
    vector<Trie*> children;
    int ref;

    Trie()
        : children(26, nullptr)
        , ref(-1) {}

    void insert(const string& w, int ref) {
        Trie* node = this;
        for (char c : w) {
            c -= 'a';
            if (!node->children[c]) {
                node->children[c] = new Trie();
            }
            node = node->children[c];
        }
        node->ref = ref;
    }
};

class Solution {
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        Trie* tree = new Trie();
        for (int i = 0; i < words.size(); ++i) {
            tree->insert(words[i], i);
        }
        vector<string> ans;
        int m = board.size(), n = board[0].size();

        function<void(Trie*, int, int)> dfs = [&](Trie* node, int i, int j) {
            int idx = board[i][j] - 'a';
            if (!node->children[idx]) {
                return;
            }
            node = node->children[idx];
            if (node->ref != -1) {
                ans.emplace_back(words[node->ref]);
                node->ref = -1;
            }
            int dirs[5] = {-1, 0, 1, 0, -1};
            char c = board[i][j];
            board[i][j] = '#';
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#') {
                    dfs(node, x, y);
                }
            }
            board[i][j] = c;
        };

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                dfs(tree, i, j);
            }
        }
        return ans;
    }
};