<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{
public:
    map<char, TrieNode*> children;
    bool isWord = false;
    int refs = 0;
    
    void addWord(string &word){

        TrieNode* cur = this;
        cur->refs++;

        for(char &c : word){
            if(cur->children.count(c) == 0)
                cur->children[c] = new TrieNode();
            cur = cur->children[c];
            cur->refs++;
        }
        cur->isWord = true;
    }

    void removeWord(string word){
        TrieNode* cur = this;
        cur->refs--;

        for(char &c : word){
            if(cur->children.count(c)){
                cur = cur->children[c];
                cur->refs--;
            }
        }
    }
    
};


class Solution {

int n,m;
vector<int> dx = {1, 0, -1, 0};
vector<int> dy = {0, 1, 0, -1};


bool inBound(int x, int y){
    if(x>=0 && x<n && y>=0 && y<m){
        return true;
    }
    return false;
}

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

        unordered_set<string> res;
        unordered_set<string> vis;

        TrieNode* root = new TrieNode();

        for(auto &word : words){
            root->addWord(word);
        }

        for(int r = 0; r<n; r++){
            for(int c = 0; c<m; c++){
                dfs(r,c,root, "", 
                    board, res, vis, root);
            }
        }

        return vector<string>(res.begin(), res.end());
    }

private:

    string stringify(int s1, int s2){
        return to_string(s1) + ',' + to_string(s2);
    }
    
    void dfs(int r, int c, TrieNode* node, string word,
    vector<vector<char>>& board, unordered_set<string> &res, 
    unordered_set<string> &vis, TrieNode* root )
    {
        if(!inBound(r,c) 
            || node->children.count(board[r][c]) == 0
            || node->children[board[r][c]]->refs <1
            || vis.count( stringify(r,c) ) > 0)
        {
                return;
        }

        vis.insert(stringify(r,c));

        node = node->children[board[r][c]];

        word += board[r][c];

        if(node->isWord){
            node->isWord = false;
            res.insert(word);
            root->removeWord(word);
        }

        for(int i = 0; i<4; i++){
            int x = r + dx[i];
            int y = c + dy[i];

            dfs(x,y, node,word, board, res, vis, root);

        }
        vis.erase(stringify(r,c));
    }

};
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