<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 {
public:
    vector<int>dx={0,0,-1,1};
    vector<int>dy={-1,1,0,0};
    int n,m;
    bool solve(int &i,int &j,vector<vector<char>>& board, string &word,string &s,set<pair<int,int>>&visited){
        if(word==s)return true;//this also prevent us from going out of bound
        if(s.size()>word.size()||visited.size()==m*n)return false;
        for(int k=0;k<4;k++){
            int ni=i+dx[k];
            int nj=j+dy[k];
            //we try to prune if we can do it if we go our of bound or not found th next char whihc is needed
            if(ni<0||nj<0||ni>=m||nj>=n||visited.find({ni,nj})!=visited.end()||board[ni][nj]!=word[s.size()]) continue;
            s+=word[s.size()];//adding the 
            visited.insert({ni,nj});
            bool found=solve(ni,nj,board,word,s,visited);
            visited.erase({ni,nj});
            s=s.substr(0,s.size()-1);//removing the char
            if(found)return true;
        }
        return false;
    }
    bool exist(vector<vector<char>>& board, string word) {
         m=board.size();
         n=board[0].size();
         reverse(word.begin(),word.end());
        char beg=word[0];
        set<pair<int,int>>visited;
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                if(board[i][j]!=beg) continue;
                string s="";
                s+=beg;
                visited.insert({i,j});
                bool found=solve(i,j,board,word,s,visited);
                visited.erase({i,j});
                if(found)return true;
            }
        }
        return false;
    }
};
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