<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 stops=0;
        int dx[4]={-1,1,0,0};
        int dy[4]={0,0,-1,1};

        unordered_set<string> visited;
        queue<vector<int>> q;
        int total=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]=='@')
                {
                    string s=to_string(i)+" "+to_string(j)+" "+to_string(0);
                    visited.insert(s);
                    q.push({i,j,0});
                }
                if(grid[i][j]>='a' && grid[i][j]<='f')
                total++;
            }
        }
        while(!q.empty()){
            int sz=q.size();
            while(sz--){
            int x,y,k;
            vector<int> temp=q.front();
            q.pop();
            x=temp[0],y=temp[1],k=temp[2];
            if(__builtin_popcount(k)==total)return stops;
            for(int i=0;i<4;i++){
                int x1,y1,k1;
                x1=x+dx[i],y1=y+dy[i],k1=k;
                
                if(x1<0 || x1>=n || y1<0 || y1>=m || grid[x1][y1]=='#')continue;
                if(grid[x1][y1]>='a' && grid[x1][y1]<='f')
                k1|=1<<(grid[x1][y1]-'a');
                string tmp=to_string(x1)+" "+to_string(y1)+" "+to_string(k);
                if(visited.find(tmp)==visited.end()){
                    if(grid[x1][y1]>='A' && grid[x1][y1]<='F' && !(k1&(1<<(grid[x1][y1]-'A'))))
                    continue;
                    //cout<<x1<<","<<y1<<"->"<<k1<<endl;
                    q.push({x1,y1,k1});
                    visited.insert(tmp);
                }
            }}
            //cout<<stops<<endl;
            stops++;
        }
        return -1;
    }
};
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