<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 {
private:
    
    unordered_set<int> used;
    
    bool backtrack(int i, int j,int k,string word, vector<vector<char>>& board){
        if(k==word.size()) return true;
        
        if(used.find(i*board[0].size()+j)!=used.end()) return false;
        
       
        
        if(i < 0 ||i >= board.size()) return false;
        if(j < 0 || j >= board[0].size()) return false;

        if(board[i][j]!=word[k]) return false;
        
        used.insert(i*board[0].size()+j);
        
        for(int l = -1; l <=1; l++){
            for(int p = -1; p <= 1; p++){
                if(l!=0 && p!=0) continue;
                int currI = i + l;
                int currJ = j + p;
                
                
                if(backtrack(currI,currJ,k+1,word,board))return true;
                 
            }
        }
        
        used.erase(i*board[0].size()+j);
        return false;
        
    }
    
    bool testIfAllLettersExists(vector<vector<char>>& board, string word){
        unordered_set<char> letters;
        for(int i = 0; i < word.size(); i++)
            letters.insert(word[i]);
        
        for(int i = 0 ;i < board.size(); i++)
            for(int j = 0; j < board[0].size(); j++)
            {
                if(letters.find(board[i][j])!=letters.end())
                {
                    letters.erase(board[i][j]);
                    if(letters.size()==0) return true;
                }
            }
        
        return false;
    }
    
public:
    
    bool exist(vector<vector<char>>& board, string word) {
        if(!testIfAllLettersExists(board,word)) return false;
        
        for(int i = 0; i < board.size();i++)
            for(int j = 0; j < board[0].size();j++)
            {
                if(backtrack(i,j,0,word,board))return true;
            }
       
        return false;
    }
};
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