<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();
        int n=grid[0].size();
        int maxkeys=-1;
        int x,y;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                char ch=grid[i][j];
                if(ch=='@'){
                    x=i;
                    y=j;
                }
                else if(ch>='a' && ch<='f'){
                    maxkeys=max(maxkeys,ch-'a'+1);
                }
            }
        }
        int ans=0;
        int dir[][2]={{0,1},{0,-1},{-1,0},{1,0}};
        vector<int> start={0,x,y};
        queue<vector<int>> q;
        q.push(start);
        unordered_set<string> visited;
        visited.insert(to_string(0) + " " + to_string(x) + " " + to_string(y));

        while(!q.empty()){
            int size=q.size();
            while(size--){
                auto curr=q.front();
                q.pop();
                int x = curr[1];
                int y = curr[2];

                if(curr[0] == ((1<<maxkeys)-1)){
                    return ans;
                }


                for(int k=0;k<4;k++){
                    int nx = x + dir[k][0];
                    int ny = y + dir[k][1];
                    int keys=curr[0];

                    if(nx>=0 && ny>=0 && nx<m && ny<n){
                        char ch = grid[nx][ny];

                        if(ch=='#'){
                            continue;
                        }
                        if(ch>='a' && ch<='f'){
                            keys |= 1<<(ch-'a');
                        }
                        if(ch>='A' && ch<='F' && (keys & (1<<(ch-'A')))==0){
                            continue;
                        }
                        if(!visited.count(to_string(keys)+" "+to_string(nx)+" "+to_string(ny))){
                            visited.insert(to_string(keys)+" "+to_string(nx)+" "+to_string(ny));
                            q.push({keys,nx,ny});
                        }
                    }
                }
            }
            ans++;
        }

        return -1;
    }
};
48173