<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:
    vector<vector<bool>> vis;
    bool helper(vector<pair<int, int>> list, vector<vector<char>>& board, string word, int ind){
        if(ind == word.size()){
            return true;
        }
        for(int i = 0; i < list.size(); i++){
            if(board[list[i].first][list[i].second] == word[ind]){
                vector<pair<int, int>> list1;
                if(list[i].first - 1 >= 0 && !vis[list[i].first - 1][list[i].second]){
                    list1.push_back({list[i].first - 1, list[i].second});
                }
                if(list[i].first + 1 < board.size() && !vis[list[i].first + 1][list[i].second]){
                    list1.push_back({list[i].first + 1, list[i].second});
                }
                if(list[i].second - 1 >= 0 && !vis[list[i].first][list[i].second - 1]){
                    list1.push_back({list[i].first, list[i].second - 1});
                }
                if(list[i].second + 1 < board[0].size() && !vis[list[i].first][list[i].second + 1]){
                    list1.push_back({list[i].first, list[i].second + 1});
                }
                    // cout << "Here to find " << endl;
                    // for(int i = 0; i < list1.size(); i++){
                    //     cout << "[" << list1[i].first << "][" << list1[i].second << "]" << endl;
                    // }
                    // cout << "Current index: " << ind + 1 << endl;
                    // cout << "From [" << list[i].first << "][" << list[i].second << "]" << endl;
                vis[list[i].first][list[i].second] = true;
                if(helper(list1, board, word, ind + 1)){
                    return true;
                }
                else{
                    vis[list[i].first][list[i].second] = false;
                }
            }
        }
        return false;
    }
    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size();
        int n = board[0].size();
        vis.assign(m, vector<bool>(n, false));
        vector<pair<int, int>> list;
        for(int k = 0; k < word.size(); k++){
            bool f = false;
            for(int i = 0; i < m; i++){
                for(int j = 0; j < n; j++){
                    if(board[i][j] == word[k]){
                        f = true;
                        break;
                    }
                }
            }
            if(!f){
                return false;
            }
        }
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                list.push_back({i, j});
            }
        }
        return helper(list, board, word, 0);
    }
};
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