<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 maxkey = 0,maxlock = 0;
        int startx = -1,starty = -1;
        for(int i = 0;i<n;i++){
            for(int j = 0;j<m;j++){
                if(grid[i][j] >= 'a' && grid[i][j] <= 'z'){
                    maxkey = max(maxkey,grid[i][j] - 'a');
                }
                if(grid[i][j] == '@'){
                    startx = i,starty = j;
                }
            }
        }
        int dr[4] = {1,-1,0,0};
        int dc[4] = {0,0,1,-1};
        maxkey++;
        int vis[1LL<<maxkey][n][m];
        memset(vis,0,sizeof(vis));
        queue<vector<int>> q;
        q.push({startx,starty,0,0}); // r,c,dis,keymask
        while(!q.empty()){
            int r = q.front()[0];
            int c = q.front()[1];
            int dis = q.front()[2];
            int mask = q.front()[3];
            q.pop();
            if(vis[mask][r][c]) continue;
            vis[mask][r][c] = 1;
            //cout << r << " " << c << " " << dis << " " << mask << endl;
            if(mask == (1<<maxkey) - 1) return dis;
            for(int i = 0;i<4;i++){
                int nr = r + dr[i];
                int nc = c + dc[i];
                if(nr < n && nc < m && nr >= 0 && nc >= 0 && grid[nr][nc] != '#' && !vis[mask][nr][nc]){
                    if(grid[nr][nc] >= 'A' && grid[nr][nc] <= 'Z' && (mask & (1<<(grid[nr][nc]-'A')))){
                        q.push({nr,nc,dis+1,mask});
                    }
                    else if(grid[nr][nc] == '.' || grid[nr][nc] == '@'){
                        q.push({nr,nc,dis+1,mask});
                    }
                    else if(grid[nr][nc] >= 'a' && grid[nr][nc] <= 'z'){
                        if((!(mask & (1<<(grid[nr][nc]-'a'))))) q.push({nr,nc,dis+1,mask ^ (1<<(grid[nr][nc]-'a'))});
                        else q.push({nr,nc,dis+1,mask});
                    }
                }
            }
        }
        return -1;
    }
};
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