<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>

typedef struct{
    int x;
    int y;
}move_t;

class Solution {
public:

    void backtrack(int x,int y,int k,vector<vector<bool>> &V,vector<vector<char>> &board,string word,bool &esito){

        if(esito){
            return;
        }

        if(k == word.size()){
            esito = true;
            return;
        }else{
            
            vector<move_t> candidates;
            if(k == 0){
                for(int i=0;i<board.size();i++){
                    for(int j=0;j<board[0].size();j++){
                        if(board[i][j] == word[k]){
                            move_t m = {i,j};
                            candidates.push_back(m);
                        }
                    }
                }
            }else{
                
                move_t m;

                if( (x+1) <= (board.size()-1) && V[x+1][y] != true && word[k] == board[x+1][y]){
                    m.x = x+1;
                    m.y = y;
                    candidates.push_back(m);
                }

                if( (x-1) >= 0 && V[x-1][y] != true && word[k] == board[x-1][y]){
                    m.x = x-1;
                    m.y = y;
                    candidates.push_back(m);
                }

                if( (y-1) >= 0 && V[x][y-1] != true && word[k] == board[x][y-1]){
                    m.x = x;
                    m.y = y-1;
                    candidates.push_back(m);
                }

                if( (y+1) <= (board[0].size()-1) && V[x][y+1] != true && word[k] == board[x][y+1]){
                    m.x = x;
                    m.y = y+1;
                    candidates.push_back(m);
                }
            }

            for(int i=0;i<candidates.size();i++){
                x = candidates[i].x;
                y = candidates[i].y;
                V[x][y] = true;
                backtrack(x,y,k+1,V,board,word,esito);
                V[x][y] = false;
            }
        }
    }

    bool exist(vector<vector<char>>& board, string word) {
        
        bool esito = false;

        vector<vector<bool>> V(board.size(),vector<bool>(board[0].size(),false));

        backtrack(0,0,0,V,board,word,esito);

        return esito;
    }
};
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