<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 {
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        end.resize(30001, "");
        buildTrie(words);
        vector<string> res;
        int n = board.size(), m = board[0].size();
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                dfs(board, i, j, 1, res);
            }
        }

        return res;
    }

    // start a dfs on board position (@i, @j) and from level @t of the Trie
    // update the collected string in @res
    // return: number of strings being collected (for pruning)
    int dfs(vector<vector<char>>& board, int i, int j, int t, vector<string>& res){
        // check if the position is out of board or searched along dfs path
        if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || board[i][j] == 0){
            return 0;
        }

        // check if there is definitely no more strings to collect at this point path
        char tmp = board[i][j]; // char at this point
        int path = tmp - 'a';
        t = tree[t][path];
        if (t == 0 || pass[t] == 0) { // no prefix exists on this level of Trie
            return 0;
        }

        // there are still some strings might could be collected along this path
        int fix = 0; // # of strings collected along this path
        if (end[t].size() > 0){
            fix++;
            res.push_back(end[t]);
            end[t] = "";
        }

        // dfs to adjacent blocks
        board[i][j] = 0;
        fix += dfs(board, i, j+1, t, res);
        fix += dfs(board, i, j-1, t, res);
        fix += dfs(board, i+1, j, t, res);
        fix += dfs(board, i-1, j, t, res);
        board[i][j] = tmp;

        // update pass (the strings remained AND with this prefix)
        pass[t] -= fix;

        return fix;
    }

    void buildTrie(vector<string>& words){
        for (const string& word: words){
            int node = 1;
            pass[node]++;
            int n = word.size();
            for (int i = 0; i < n; i++){
                int path = word[i] - 'a';
                if (tree[node][path] == 0) {
                    tree[node][path] = ++cnt; 
                }
                node = tree[node][path]; // next node
                pass[node]++;
            }
            end[node] = word;
        }
    }

    int tree[30001][26] = {0};
    int pass[30001] = {0};
    vector<string> end;
    int cnt = 1;
};
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