<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, string word, int i, int j, int n){
        
        if(n==word.size()) return true;
        if(i<0 || j<0 || i>=board.size() || j>=board[0].size()) return false;
        
        if(board[i][j] == '0') return false;
        if(board[i][j] != word[n]) return false;
       
        board[i][j] = '0';
        cout<<i<<" "<<j<<endl;
        if(solve(board,word,i-1,j,n+1) || solve(board,word,i+1,j,n+1) || solve(board,word,i,j-1 ,        n+1) || solve(board,word,i,j+1,n+1)) return true;
        return false;
    }
    bool exist(vector<vector<char>>& board, string word) {
        vector<bool> ans;
        unordered_set<char> charSet;
        map<char, int> fre;
        for (char c : word) {
            fre[c]++;
        }
        if(fre[word[word.size()-1]] < fre[word[0]]) reverse(word.begin(),word.end());
        cout<<word;
        for (const auto& row : board) {
            for (char c : row) {
                charSet.insert(c);
            }
        }
        for (int i = 0;i < word.size();i++) {
            char c = word[i];
            if (charSet.find(c) == charSet.end()) {
                return false; 
            }
        }
       for(int i = 0;i < board.size();++i){
            for(int j = 0;j<board[0].size();++j){
                
            }
       }
        for(int i = 0;i < board.size();++i){
            for(int j = 0;j<board[0].size();++j){
                if(board[i][j] == word[0]){
                    //cout<<i<< " " <<j<<endl;
                    if(solve(board,word,i,j,0)) return true;
                }
            }
        }
        return false;
    }
};
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