<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:
    int index(int i, int j, int m) {
        return i * m + j;
    }
    bool busy(uint64_t mask, int i, int j, int m) {
        int idx = index(i, j, m);
        return (mask & (1ULL << idx)) > 0;
    }
    uint64_t set_bit(uint64_t mask, int i, int j, int m) {
        int idx = index(i, j, m);
        return mask | (1ULL << idx);
    }
    void check(queue<pair<int, uint64_t>>& q, const vector<vector<char>>& board, const string& word, int i, int j) {
        int n = board.size();
        int m = board[0].size();
        auto p = q.front();
        uint64_t mask = p.second;
        int bits = __popcount(mask);
        bool inside = i >= 0 && i < n && j >= 0 && j < m;
        if (inside && !busy(mask, i, j, m) && board[i][j] == word[bits]) {
            int new_idx = index(i, j, m);
            uint64_t new_mask = set_bit(mask, i, j, m);
            q.push(make_pair(new_idx, new_mask));
        }
    }
    bool exist(vector<vector<char>>& board, string word) {
        int n = board.size();
        int m = board[0].size();
        int L = word.size();
        queue<pair<int, uint64_t>> q;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (board[i][j] == word[0]) {
                    int idx = index(i, j, m);
                    q.push(make_pair(idx, 1ULL << idx));
                }
            }
        }
        while (!q.empty()) {
            auto p = q.front();
            uint64_t mask = p.second;
            int bits = __popcount(mask);
            if (bits == L) {
                return true;
            }
            int i = p.first / m;
            int j = p.first % m;
            check(q, board, word, i - 1, j);
            check(q, board, word, i + 1, j);
            check(q, board, word, i, j + 1);
            check(q, board, word, i, j - 1);
            q.pop();
        }
        return false;
    }
};
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