<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:

    struct Trie {
        Trie* children[26];
        bool isEnd;
        int num;
        Trie() {
            isEnd = false;
            num = 0;
            for(int i = 0; i < 26; ++i)
                children[i] = nullptr;
        }
    };

    Trie* root;

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        root = new Trie();
        //Add words
        for(auto word: words) {
            Trie* curr = root;
            for(auto ch: word) {
                if(curr->children[ch-'a'] == nullptr)
                    curr->children[ch-'a'] = new Trie();
                curr = curr->children[ch - 'a'];
                ++(curr->num);
            }
            curr->isEnd = true;
        }
        vector<string> answer;
        //Backtrack through board
        int m = board.size();
        int n = board[0].size();
        unordered_set<int> visited;
        unordered_set<string> seen;
        for(int i = 0; i < m; ++i) {
            for(int j = 0; j < n; ++j) {
                if(root->children[board[i][j] - 'a'] != nullptr && root->children[board[i][j] - 'a']->num > 0) {
                    string temp = "";
                    temp.push_back(board[i][j]);
                    visited.clear();
                    visited.insert(i * n + j);
                    backtrack(board, i, j, m, n, answer, root->children[board[i][j] - 'a'], temp, visited); 
                }
            }
        }
        return answer;
    }

    int rows[4] = {-1, 1, 0, 0};
    int cols[4] = {0, 0, -1, 1};
    void backtrack(vector<vector<char>>& board, int i, int j, int m, int n, vector<string>& answer, Trie* curr, 
        string& curString, unordered_set<int>& visited) {
        if(curr->num <= 0)
            return;

        if(curr->isEnd) {
            //Subtract along path
            Trie* curr = root;
            for(auto ch: curString) {
                curr = curr->children[ch - 'a'];
                --(curr->num);
            }
            curr->isEnd = false;
            answer.push_back(curString);
        }

        for(int k = 0; k < 4; ++k) {
            int r = i + rows[k];
            int c = j + cols[k];

            if(!(r >= 0 && r < m && c >= 0 && c < n) || visited.count(r * n + c) > 0)
                continue;
            
            if(curr->children[board[r][c] -'a'] != nullptr) {
                curString.push_back(board[r][c]);
                visited.insert(r * n + c);
                backtrack(board, r, c, m, n, answer, curr->children[board[r][c] -'a'], curString, visited);
                curString.pop_back();
                visited.erase(r * n + c);
            }
        }
    }
};
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