<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:
    vector<vector<int>>directions{{1,0},{-1,0},{0,1},{0,-1}};
    int shortestPathAllKeys(vector<string>& grid) {
        int m=grid.size();
        int n=grid[0].size();
        queue<vector<int>>q;
        int count=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='@'){
                    q.push({i,j,0,0});
                    

                }
                if(grid[i][j]>='a' && grid[i][j]<='f'){
                    count++;
                }
            }
        }
        int s=pow(2,count)-1;
        vector<vector<vector<int>>>visited(m,vector<vector<int>>(n,vector<int>(s+1,0)));
        // memset(visited,0,sizeof(visited));
        while(!q.empty()){
            vector<int>node=q.front();
            q.pop();
            int row=node[0];
            int col=node[1];
            int c=node[2];
            int score=node[3];
            if(score==s) return c;
            for(auto it:directions){
                int x=row+it[0];
                int y=col+it[1];
                if(x>=0 && y>=0 && x<m && y<n && grid[x][y]!='#'){
                    if(grid[x][y]>='A' && grid[x][y]<='F' ){
                        if(visited[x][y][score]==0 && (score & (1<<(grid[x][y]-'A')))!=0){
                            visited[x][y][score]=1;
                            q.push({x,y,c+1,score});
                        }
                    } 
                    else if(grid[x][y]>='a' && grid[x][y]<='f'){
                        if(visited[x][y][score]==0 ){
                            visited[x][y][score]=1;
                            int newscore=score | (1<< (grid[x][y]-'a'));
                            q.push({x,y,c+1,newscore});
                        }

                    }
                    else{
                        if(visited[x][y][score]==0){
                            visited[x][y][score]=1;
                            q.push({x,y,c+1,score});

                        }
                        
                    }

                }
            }

        }
        return -1;
        


        
    }
};
59010