<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:
    bool fun(int i,int j,string s1,string word,vector<vector<int>>&vis,vector<vector<char>>& board){
        if(s1==word||s1+board[i][j]==word)return true;
        if(s1.length()>=word.length())return false;
        vector<int>x1{0,0,1,-1},y1{1,-1,0,0};
        for(int k=0;k<4;k++){
            int nx=i+x1[k],ny=j+y1[k];
            if(nx>=0&&ny>=0&&nx<board.size()&&ny<board[0].size()&&!vis[nx][ny])
            {
                if(word[s1.size()]==board[i][j]){
                vis[nx][ny]=1;
                if(fun(nx,ny,s1+board[i][j],word,vis,board))return true;
                vis[nx][ny]=0;
                }
            }
        }
        
        return false;
    }
    bool exist(vector<vector<char>>& board, string word) {
        vector<vector<int>> vis(board.size(),vector<int>(board[0].size(),0));
        map<char,int>mp;
        map<char,int>mp1;
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                int flg=true;
                mp[board[i][j]]++;
               for(int k=0;k<word.size();k++){
                mp1[word[k]]++;
                if(board[i][j]==word[k])
                flg=false;
               }
               if(flg)vis[i][j]=-1;
            }
        }
        for(int k=0;k<word.size();k++){
                if(mp[word[k]]==0)
                return false;
               }
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                vis[i][j]=1;
                if(fun(i,j,"",word,vis,board))return true;
                vis[i][j]=0;
            }
        }
        return false;
    }
};
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