<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<int>> delta;
    bool exist(vector<vector<char>>& board, string word) {
        
        int row = board.size();
        int col = board[0].size();
        int n = word.size();
        if(n==0){
            return true;
        }
        if(n>row*col){
            return false;
        }
        //check the frequency
        vector<char> boardFreq(128,0);
        vector<char> stringFreq(128,0);
        for(auto c:word){
            stringFreq[c]++;
        }
        for(int i=0; i<row; i++){
            for(int j=0; j<col; j++){
                boardFreq[board[i][j]]++;
            }
        }
        for(int i=0; i<128; i++){
            if(stringFreq[i]>boardFreq[i]){
                return false;
            }
        }
        //dfs
        vector<bool> usedLetterIndex(row*col,false);
        delta.push_back({0,1});
        delta.push_back({1,0});
        delta.push_back({0,-1});
        delta.push_back({-1,0});

        for(int i=0; i<row; i++){
            for(int j=0; j<col; j++){
                if (dfs(board,usedLetterIndex,i,j,word,0)){
                    return true;
                };
            }
        }
        return false;
    }
    bool dfs(vector<vector<char>>& board,vector<bool>& usedLetterIndex, int i,int j,string& word,int wordIndex){
        if(wordIndex==word.size()){
            return true;
        }
        int row = board.size();
        int col = board[0].size();
        if(i<0 ||i>=row ||j<0||j>=col){
            return false;
        }
        if(board[i][j]!=word[wordIndex]){
            return false;
        }
        int usedIndex = col*i+j;

        if(usedLetterIndex[usedIndex]){
            return false;
        }

        usedLetterIndex[usedIndex]=true;
        bool result=false;
        for(auto v:delta){
            int dx = v[0];
            int dy = v[1];
            result= (result || dfs(board,usedLetterIndex,i+dx,j+dy,word,wordIndex+1));
            if(result){
                break;
            }
        }
        usedLetterIndex[usedIndex]=false;
        return result;
    }
};
160218