<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 myfunc(vector<vector<char>> board, string word,int r=0,int c=0) {
        //base condition for word found
        if(word.size()==0) return true;
        int R,C;
        R=board.size();
        C=board[0].size();
        //base conditions for word not found
        if(r>=R || r<0 || c>=C || c<0) return false;
        if(board[r][c]== '-1' || board[r][c]!=word[0]) return false;

        //checking way further 
        board[r][c]='-1';
        string new_word= word.substr(1,word.size()-1);
        if(word.size()==1 ) return true;


        if( myfunc(board,new_word,r,c-1)) return true;
        if( myfunc(board,new_word,r,c+1)) return true;
        if(myfunc(board,new_word,r-1,c)) return true;
        if( myfunc(board,new_word,r+1,c)) return true;

        
                
        return false;        
    }   


    bool exist(vector<vector<char>>& board, string word) {
        int i,j,r,c,R,C;
        int freq[200]={0};
        R=board.size();
        C=board[0].size();

        // frequency checking
        for(i=0;i<R;i++)
            for(j=0;j<C;j++)
                freq[int(board[i][j])]++;

        // reversing word if occurence of last letter is less than the first letter
        if(freq[int(word[word.size()-1])] < freq[int(word[0])])
            reverse(word.begin(),word.end());
        
        for(i=0;i<word.size();i++)
            freq[int(word[i])]--;

        // if less occurences of letter are in the board the word cant be formed
        for(i=0;i<200;i++)
            if(freq[i]<0) return false;


        for(r=0;r<R;r++)
            for(c=0;c<C;c++)
                if(myfunc(board,word,r,c)) return true;


        return false;  
    }
};
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