<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 m = grid.size(), n = grid[0].size();
        int num_keys = 0;
        int sx = -1, sy = -1;
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if ((grid[i][j] - 'a') >=0 && (grid[i][j] - 'a') < 26)  num_keys++;
                else if (grid[i][j] == '@') {
                    sx = i, sy = j;
                }
            }
        }
        queue<vector<int>>q;
        q.push({sx, sy, 0});
        int steps = 0;
        set<vector<int>>v;
        vector<pair<int, int>>dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        while (!q.empty()) {
            int sz = q.size(); 
            while (sz--) {
                int x = q.front()[0], y = q.front()[1], mask = q.front()[2];
                q.pop();
                if (mask == ((1 << num_keys) - 1))    return steps;
                for (auto [dx, dy]: dirs) {
                    int nx = x + dx, ny = y + dy;
                    if (nx < 0 || nx >= m || ny < 0 || ny >= n || grid[nx][ny] == '#')  continue;
                    int new_mask = mask;
                    if (grid[nx][ny] >= 'a' && grid[nx][ny] <= 'z') {
                        int bit = grid[nx][ny] - 'a';
                        new_mask = new_mask | 1 << bit;
                    } else if (grid[nx][ny] >= 'A' && grid[nx][ny] <= 'Z') {
                        int bit = grid[nx][ny] - 'A';
                        if ((new_mask >> bit) & 1);
                        else    continue;
                    }
                    if (v.find({nx, ny, new_mask}) != v.end())    continue;
                    else    v.insert({nx, ny, new_mask});
                    q.push({nx, ny, new_mask});
                }
            }
            steps++;
        }
        return -1;
    }
};
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