<p>You are given an <code>m x n</code> grid <code>grid</code> where:</p>

<ul>
	<li><code>&#39;.&#39;</code> is an empty cell.</li>
	<li><code>&#39;#&#39;</code> is a wall.</li>
	<li><code>&#39;@&#39;</code> is the starting point.</li>
	<li>Lowercase letters represent keys.</li>
	<li>Uppercase letters represent locks.</li>
</ul>

<p>You start at the starting point and one move consists of walking one space in one of the four cardinal directions. You cannot walk outside the grid, or walk into a wall.</p>

<p>If you walk over a key, you can pick it up and you cannot walk over a lock unless you have its corresponding key.</p>

<p>For some <code><font face="monospace">1 &lt;= k &lt;= 6</font></code>, there is exactly one lowercase and one uppercase letter of the first <code>k</code> letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.</p>

<p>Return <em>the lowest number of moves to acquire all keys</em>. If it is impossible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-keys2.jpg" style="width: 404px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@.a..&quot;,&quot;###.#&quot;,&quot;b.A.B&quot;]
<strong>Output:</strong> 8
<strong>Explanation:</strong> Note that the goal is to obtain all the keys not to open all the locks.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-key2.jpg" style="width: 404px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@..aA&quot;,&quot;..B#.&quot;,&quot;....b&quot;]
<strong>Output:</strong> 6
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-keys3.jpg" style="width: 244px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [&quot;@Aa&quot;]
<strong>Output:</strong> -1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either an English letter, <code>&#39;.&#39;</code>, <code>&#39;#&#39;</code>, or <code>&#39;@&#39;</code>.&nbsp;</li>
	<li>There is exactly one&nbsp;<code>&#39;@&#39;</code>&nbsp;in the grid.</li>
	<li>The number of keys in the grid is in the range <code>[1, 6]</code>.</li>
	<li>Each key in the grid is <strong>unique</strong>.</li>
	<li>Each key in the grid has a matching lock.</li>
</ul>

class Solution {
public:
    int shortestPathAllKeys(vector<string>& grid) {
        int n = grid.size(), m = grid[0].size();
        int k = 0;
        int si = 0, sj = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(grid[i][j] >= 'a' && grid[i][j] <= 'z')
                    k = max(k, grid[i][j] - 'a');
                if(grid[i][j] == '@')
                {
                    si = i;
                    sj = j;
                }
            }
        }
        int allKeys = (1 << (k + 1)) - 1;
        vector<vector<vector<int>>> dis(n, vector<vector<int>>(m, vector<int>(allKeys + 1, INT_MAX)));
        vector<vector<int>> dirs{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        multiset<vector<int>> pq;
        pq.insert({0, si, sj, 0});
        dis[si][sj][0] = 0;
        while(!pq.empty())
        {
            vector<int> v = *pq.begin();
            pq.erase(pq.begin());
            int d = v[0];
            int ci = v[1], cj = v[2];
            int keys = v[3];
            cout << keys << "\n";
            if(keys == allKeys)
                return d;
            for(vector<int> &dir: dirs)
            {
                int ni = ci + dir[0], nj = cj + dir[1];
                if(ni < 0 || ni >= n || nj < 0 || nj >= m)
                    continue;
                if(grid[ni][nj] == '#') // wall
                    continue;
                if(grid[ni][nj] >= 'A' && grid[ni][nj] <= 'Z') // lock
                {
                    if((keys & (1 << (grid[ni][nj] - 'A'))) == 0)
                        continue;
                }
                int nkey = keys;
                if(grid[ni][nj] >= 'a' && grid[ni][nj] <= 'z')
                    nkey |= (1 << (grid[ni][nj] - 'a'));
                if(dis[ni][nj][nkey] <= d + 1)
                    continue;
                dis[ni][nj][nkey] = d + 1;
                pq.insert({d + 1, ni, nj, nkey});
            }
        }
        return -1;
    }
};
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