<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:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        n = len(grid)
        m = len(grid[0])
        lower = {"a","b","c","d","e","f"}
        upper = {"A","B","C","D","E","F"}
        k = 0
        q = deque()
        vis = set()
        for i in range(n):
            for j in range(m):
                if grid[i][j] == "@":
                    q.append([i,j,0,0])
                    vis.add((i,j,0,0))
                elif grid[i][j] in lower:
                    k+=1
        dir_x = [0,0,-1,1]
        dir_y = [-1,1,0,0]
        
        
        while q:
            x,y,distance,mask = q.popleft()
            # print(x,y,distance,mask)
            if mask == 2**k-1:
                return distance
            
            for d in range(4):
                new_x,new_y = x+dir_x[d],y+dir_y[d]
                
                if new_x>=0 and new_x<n and new_y>=0 and new_y<m and grid[new_x][new_y]!="#":
                    ch = grid[new_x][new_y]
                    if ch in upper and mask&(1<<(ord(ch)-ord('A'))) and (new_x,new_y,mask) not in vis:
                        vis.add((new_x,new_y,mask))
                        q.append([new_x,new_y,distance+1,mask])
                    elif ch in lower:
                        newKey = mask|(1<<(ord(ch)-ord('a')))
                        if (new_x,new_y,newKey) not in vis:
                            vis.add((new_x,new_y,newKey))
                            q.append([new_x,new_y,distance+1,newKey])
                    elif (ch == "." or ch == "@") and (new_x,new_y,mask) not in vis:
                        vis.add((new_x,new_y,mask))
                        q.append([new_x,new_y,distance+1,mask])
                    
           
        return -1

        
20481