<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: unordered_set<char> myset;
public:
    bool exist(vector<vector<char>>& board, string word) {
        const int m = board.size();
        const int n = board[0].size();
        vector<int> temp;
        for(int i = 0;i<m;i++){
            for(int j = 0;j<n;j++){
                temp.push_back(i*n + j);
                myset.insert(board[i][j]);
            }
        }
        for(char i : word){
            if(myset.find(i) == myset.end()) return false;
        }
        return solve(board,word,temp);
    }

    bool solve(vector<vector<char>>& board,string word,vector<int> start){
        if(word.size() == 0){ return true;}
        const int m = board.size();
        const int n = board[0].size();

        for(int i: start){
            int row = i/n;
            int column = i%n;

            if(word[0] == board[row][column]){
                vector<int> temp;
                if(row > 0 && board[row-1][column]!='#') temp.push_back((row-1)*n + column);
                if(row < (m-1) && board[row+1][column] != '#') temp.push_back((row+1)*n + column);
                if(column > 0 && board[row][column - 1] != '#') temp.push_back(row*n + column - 1);
                if(column < (n-1) && board[row][column + 1] != '#') temp.push_back(row*n + column + 1);
                board[row][column] = '#';
                bool ans = solve(board,word.substr(1),temp);
                board[row][column] = word[0];
                if(ans) return true;
            }
        }

        return false;
    }
};
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