<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 {
    const int dx[4] = {-1, 1, 0, 0};
    const int dy[4] = {0, 0, -1, 1};
    bool islower(char x) { return (x >= 'a' && x <= 'z');}
    bool isvalid(int x, int y, int rows, int cols) {
        return (x >= 0 && x < rows && y >= 0 && y < cols);
    }
    bool isupper(char x) { return (x >= 'A' && x <= 'Z'); }
    int setbits(int mask) {
        int res = 0;
        while (mask) {
            res += (mask & 1);
            mask >>= 1;
        }
        return res;
    }

public:
    int shortestPathAllKeys(vector<string>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        int keys = 0;
        queue<vector<int>> q;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (islower(grid[i][j]))
                    keys++;
                if (grid[i][j] == '@')
                    q.push({i, j, 0});
            }
        }
        set<vector<int>> vis;
        int steps = 0;
        while (!q.empty()) {
            int sz = q.size();
            for (int i = 0; i < sz; i++) {
                vector<int> curr = q.front();
                q.pop();
                int x = curr[0], y = curr[1], mask = curr[2];
                if (setbits(mask) == keys)
                    return steps;
                for (int k = 0; k < 4; k++) {
                    int newx = x + dx[k], newy = y + dy[k];
                    if (!isvalid(newx,newy,rows,cols)||grid[newx][newy]=='#')continue;
                    char newchar=grid[newx][newy];
                    int newmask=mask;
                    if (islower(newchar)){
                        newmask|=(1<<newchar-'a');
                    }
                    if (vis.find({newx,newy,newmask})!=vis.end()||(isupper(newchar)&&!(mask&(1<<(newchar-'A')))))continue;
                    q.push({{newx,newy,newmask}});
                    vis.insert({newx,newy,newmask});
                }
            }steps++;
        }
        return -1;
    }
};
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