<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 a,b;
        int k=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='@'){
                    a=i;
                    b=j;
                }
                if(grid[i][j]>='a'&&grid[i][j]<='f'){
                    k++;
                }
            }
        }
        vector<int>v1={0,1,0,-1};
        vector<int>v2={1,0,-1,0};
        int state=(1<<k)-1;
        vector<vector<vector<bool>>>visited(m,vector<vector<bool>>(n,vector<bool>(state+1,0)));
        queue<vector<int>>q;
        q.push({a,b,0});
        visited[a][b][0]=1;
        int moves=0;
        while(!q.empty()){
            int c=q.size();
            for(int i=0;i<c;i++){
                vector<int>v=q.front();
                int d=v[0];
                int e=v[1];
                int f=v[2];
                if(f==state){
                    return moves; 
                }
                q.pop();
                for(int p=0;p<4;p++){
                    int x=v1[p]+d;
                    int y=v2[p]+e;
                    if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]!='#'){
                        if(grid[x][y]>='A'&&grid[x][y]<='F'){
                            int c=1<<(grid[x][y]-'A');
                            if(c&f){
                                if(visited[x][y][f]==0){
                                    q.push({x,y,f});
                                    visited[x][y][f]=1;
                                }
                            }
                        }
                        else if(grid[x][y]>='a'&&grid[x][y]<='f'){
                            int c=1<<(grid[x][y]-'a');
                            int d=c|f;
                            if(visited[x][y][d]==0){
                                q.push({x,y,d});
                                visited[x][y][d]=1;
                            }
                        }
                        else{
                            if(visited[x][y][f]==0){
                                q.push({x,y,f});
                                visited[x][y][f]=1;
                            }
                        }
                    }
                }
            }
            moves++;
        }
        return -1;
    }
};
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