<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 solve(vector<vector<char>>& board,int row,int col,int index,set<pair<int,int>> &visited,string &word){
    if(row<0 || col<0 || row>=board.size() || col>=board[0].size() || board[row][col]!=word[index] || visited.find(make_pair(row,col))!=visited.end())return false;
    if(index==word.size()-1)return true;
    visited.insert(make_pair(row,col));
    bool left=solve(board,row,col-1,index+1,visited,word);
    bool right=solve(board,row,col+1,index+1,visited,word);
    bool up=solve(board,row-1,col,index+1,visited,word);
    bool down=solve(board,row+1,col,index+1,visited,word);
    if(left||right||up||down)
        return true;
    visited.erase(make_pair(row,col));
    return false;

    }
    bool exist(vector<vector<char>>& board, string word) {
        vector<int> arr1(52,0),arr2(52,0);
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                if(board[i][j]>='a' && board[i][j]<='z')
                    arr1[(int)board[i][j]-97+26]++;
                else
                    arr1[(int)board[i][j]-65]++;
            }
        }
        for(char c:word){
            if(c>='a' && c<='z')
                arr2[int(c)-97+26]++;
            else
                arr2[int(c)-65]++;
        }
        for(int i=0;i<52;i++){
            if(arr1[i]<arr2[i])return false;
        }
        for(int i=0;i<board.size();i++){
            for(int j=0;j<board[0].size();j++){
                if(board[i][j]!=word[0])continue;
            set<pair<int,int>> st;
            if(solve(board,i,j,0,st,word))return true;
            }
        }
    return false;
    }
};
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