<p>Given an <code>m x n</code> grid of characters <code>board</code> and a string <code>word</code>, return <code>true</code> <em>if</em> <code>word</code> <em>exists in the grid</em>.</p>

<p>The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word2.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCCED&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;SEE&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/word3.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCB&quot;
<strong>Output:</strong> false
</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;= 6</code></li>
	<li><code>1 &lt;= word.length &lt;= 15</code></li>
	<li><code>board</code> and <code>word</code> consists of only lowercase and uppercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you use search pruning to make your solution faster with a larger <code>board</code>?</p>

class Solution {
private:
    
    bool recurseFindWord(vector<vector<char>>& board, string word, vector<vector<bool>>& used, int x, int y){
        if (word.size() == 0){
            return true;
        }
        if (y < 0 || y==board.size() || x<0 || x==board[0].size()){
            return false;
        }
        // cout << "rlevel " << x << " " << y << " " << used[y][x] << endl;
        if (!used[y][x] && board[y][x] == word[0]){
            used[y][x] = true;
            unordered_set<int> steps{-1,1};
            for (int x_step : steps){
                if (recurseFindWord(board, word.substr(1,word.size()-1),used,x+x_step,y)){
                    return true;
                }
            }
            for (int y_step : steps){
                if (recurseFindWord(board, word.substr(1,word.size()-1),used,x,y+y_step)){
                    return true;
                }
            }
            used[y][x] = false;
            
        }
        return false;
    }
public:
    bool exist(vector<vector<char>>& board, string word) {
        if (board.size() == 0 || board[0].size() == 0){
            return false;
        }

        // check if all letters in word are on board in the first place
        unordered_set<char> unique_letters;
        for (char ch : word){
            unique_letters.insert(ch);
        }
        for (int y=0; y<board.size(); y++){
            for (int x=0; x<board[0].size(); x++){
                if (unique_letters.find(board[y][x]) != unique_letters.end()){
                    unique_letters.erase(board[y][x]);
                    if (unique_letters.empty()){
                        break;
                    }
                }
            if (unique_letters.empty()){
                break;
            }
            }
        }
        if (!unique_letters.empty()){
            return false;
        }


        vector<vector<bool>> used(board.size(),vector<bool>(board[0].size(),false));
        for (int y=0; y<board.size(); y++){
            for (int x=0; x<board[0].size(); x++){
                // cout << x << " " << y << " " << used[y][x] << endl;
                bool res = recurseFindWord(board, word, used, x, y);
                if (res){
                    return true;
                }                
            }
        }
        return false;
        
        
    }
};
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