<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>

int row[]={0,0,1,-1};
int col[]={1,-1,0,0};
class Solution {
public:
    int shortestPathAllKeys(vector<string>& a) {
        queue <array<int,3>> q;
        int n=a.size(),m=a[0].size();
        bool vis[n][m][64];
        memset(vis,false,sizeof(vis));
        int tot=0;
        for (int i=0;i<n;i++){
            for (int j=0;j<m;j++){
                if (a[i][j]=='@'){
                    q.push({i,j,0});
                    vis[i][j][0]=true;
                }
                else if (a[i][j]>='a' && a[i][j]<='z') tot++;
            }
        }
        int ans=0;
        while (!q.empty()){
            int sz=q.size();
            for (int i=0;i<sz;i++){
                auto [x,y,mask]=q.front();
                q.pop();
            //    cout<<x<<" "<<y<<" "<<mask<<"\n";
                if (mask == ((1<<tot)-1)){
                    return ans;
                }
                for (int k=0;k<4;k++){
                    int r=x+row[k],c=y+col[k];
                    if (r>=0 && c>=0 && r<n && c<m && a[r][c]!='#'){
                        int curr=mask;
                        if (a[r][c]>='a' && a[r][c]<='z'){
                            int num=a[r][c]-'a';
                            if (!vis[r][c][mask|(1<<num)]){
                                vis[r][c][mask|(1<<num)]=true;
                                q.push({r,c,mask|(1<<num)});
                            }
                        }
                        else if (a[r][c]>='A' && a[r][c]<='Z'){
                            int req=a[r][c]-'A';
                            if (mask&(1<<req)){
                                if (!vis[r][c][mask]){
                                    vis[r][c][mask]=true;
                                    q.push({r,c,mask});
                                }
                            }
                        }
                        else{
                            if (!vis[r][c][mask]){
                                vis[r][c][mask]=true;
                                q.push({r,c,mask});
                            }
                        }
                    }
                }
            }
            ans++;
        }
        return -1;
    }
};
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