<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(), n=m?grid[0].size():0;
        if(!m || !n) return 0;
        int path=0, K=0;
        vector<int> dirs={0,-1,0,1,0};
        vector<vector<vector<bool>>> visited(m,vector<vector<bool>>(n,vector<bool>(64,0))); //at most 6 keys, using bitmap 111111
        queue<pair<int,int>> q;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='@'){
                    q.push({i*n+j,0});
                    visited[i][j][0]=1;                    
                }
                if(grid[i][j]>='A' && grid[i][j]<='F') K++;
            }
        }
        while(!q.empty()){
            int size=q.size();
            for(int i=0;i<size;i++){
                int a=q.front().first/n, b=q.front().first%n;
                int carry=q.front().second;
                q.pop();        
                if(carry==((1<<K)-1)) return path;
                for(int j=0;j<4;j++){
                    int x=a+dirs[j], y=b+dirs[j+1], k=carry;
                    if(x<0 || x>=m || y<0 || y>=n || grid[x][y]=='#') continue;
                    if(grid[x][y]>='a' && grid[x][y]<='f'){
                        k=carry|(1<<(grid[x][y]-'a'));
                    }
                    else if(grid[x][y]>='A' && grid[x][y]<='F'){
                        if(!(carry & (1<<(grid[x][y]-'A')))) continue;
                    }
                    if(!visited[x][y][k]){
                        visited[x][y][k]=1;
                        q.push({x*n+y,k});
                }                
                }
            }
            path++;
        } //Code With Avinash
        return -1;
    }
};
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