<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>

#define MAXNODES 300000

class Solution {
public:
    struct Node {
        static inline int nxt = 0;
        bool isLeaf;
        int children[26];

        Node() : isLeaf(false) {
            fill(children, children+26, 0);
        }

        int get(char c) { return children[c-'a']; }

        int set(char c) {
            children[c-'a'] = ++nxt;
            return nxt;
        }
    };
    Node trie[MAXNODES+1];

    void addWord(const string& s) {
        int curr = 0;
        for (char c : s) {
            int idx = trie[curr].get(c);
            if (idx)
                curr = idx;
            else
                curr = trie[curr].set(c);
        }
        trie[curr].isLeaf = true;
    }

    int m, n;
    vector<string> ans;

    void dfs(vector<vector<char>>& board, int r, int c, int curr, string& currString) {
        static const int dr[4] = {-1, 0, 1, 0};
        static const int dc[4] = {0, 1, 0, -1};

        int idx = trie[curr].get(board[r][c]);
        if (not idx)
            return;

        char cell = board[r][c];
        currString.push_back(cell);
        if (trie[idx].isLeaf) {
            trie[idx].isLeaf = false;
            ans.push_back(currString);
        }
        board[r][c] = '#';

        for (int i = 0; i < 4; ++i) {
            int ri = r + dr[i];
            int ci = c + dc[i];
            if (ri < 0 or ri >= m or ci < 0 or ci >= n or board[ri][ci] == '#')
                continue;
            dfs(board, ri, ci, idx, currString);
        }

        board[r][c] = cell;
        currString.pop_back();
    }

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        for (const string& word : words)
            addWord(word);
        m = board.size();
        n = board[0].size();

        string s;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                dfs(board, i, j, 0, s);
        return ans;
    }
};
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