<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 exist(vector<vector<char>>& board, string word) {

        int rows = board.size(), cols = rows? board[0].size(): 0;
        vector<pair<int,int>> dirs{{1,0},{-1,0},{0,1},{0,-1}};
        
        if(word.size() == 0|| !rows)
            return false;
        

        auto checkBound = [&rows, &cols](int r, int c){
            return r<rows && r>=0 && c<cols && c>=0;
        };
        
        stack<pair<int, pair<int,int>>> q;
        for(int  i = 0; i<rows ; i++){
            for(int j = 0; j<cols; j++){
                if(board[i][j] != word[0])
                    continue;
                
                vector<vector<bool>> visited(rows, vector<bool>(cols, false));
                q.push({i, {j, 0}});
                
                while(!q.empty()){
                    int x = q.top().first;
                    int y = q.top().second.first;
                    int idx = q.top().second.second;
                    q.pop();

                    if(idx == INT_MAX){
                        visited[x][y] = false;                 // BackTracking the wrong decision
                        continue;
                    }
                    
                    if(board[x][y] != word[idx])
                        continue;

                    if(++idx == word.size())
                        return true;
                    visited[x][y] = true;

                    for(auto dir : dirs){
                        int r = x+dir.first;
                        int c = y+dir.second;
                        if(checkBound(r, c) && !visited[r][c]){
                            q.push({r, {c, INT_MAX}});               // visualize in stack; this is for backtracing
                            q.push({r, {c, idx}});
                        }
                    }
                }
            }
        }
        
        return false;
    }
};
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