<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 {
    //Pruning: https://leetcode.com/problems/word-search/discuss/1908561/C%2B%2B%3A-How-to-prune-the-DFS-to-0ms
    
    bool solve(int idx, int x, int y, int m, int n, vector<vector<char>> &arr, string &word,                                vector<vector<int>> &vis)
    {
        if(idx == word.size())
            return true;
        
        int flag = 0;
        vector<int> dx{-1,0,1,0};
        vector<int> dy{0,1,0,-1};
        
        for(int i=0;i<4;i++)
        {
            int nx = x+dx[i];
            int ny = y+dy[i];
            if(nx>=0 and ny>=0 and nx<m and ny<n and arr[nx][ny] == word[idx] and !vis[nx][ny])
            {
                vis[nx][ny] = 1;
                if(solve(idx+1,nx, ny, m, n, arr, word, vis))
                    return true;
                vis[nx][ny] = 0;
            }
            
        }
        
        return false;

    }
public:
    bool exist(vector<vector<char>>& arr, string word) {
        
        int m = arr.size(), n = arr[0].size();
        
        //Pruning
        if(word.size() > m*n)
            return false;
        
        unordered_map<char, int> mp;
        for(int x=0;x<m;x++)
            for(int y=0;y<n;y++)
                mp[arr[x][y]]++;
        
        for(auto &a:word)
        {
            mp[a]--;
            if(mp[a]<0)
                return false;
        }
            
            
        vector<vector<int>> vis(m, vector<int>(n,0));
        int idx = 0;
        
        for(int x=0;x<m;x++)
        {
            for(int y=0;y<n;y++)
            {
                if(word[idx] == arr[x][y])
                {
                    vis[x][y] = 1;
                    if(solve(idx+1,x, y, m, n, arr, word, vis))
                        return true;
                        
                    vis[x][y] = 0;
                }
            }
        }
        
        return false;
    }
};
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