<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:
        m=len(grid)
        n=len(grid[0])

        moves=[(0,1),(0,-1),(1,0),(-1,0)]
      
        dist_keys=0
        visited=set()
        dist={}
        num_of_keys =0
        for r in range(m):
            for c in range(n):
                if grid[r][c]=='@':
                    q=[(r,c,"")]
                    visited.add((r,c,""))
                    dist[(r,c,"")]=0
                if grid[r][c] in "abcdef":
                    num_of_keys+=1

        
        while q:
            r,c,keys=q.pop(0)
            if len(keys)==num_of_keys:
                return dist[(r,c,keys)]
            for x,y in moves:
                nr = r +x
                nc = c +y
                if 0<=nr<m and 0<=nc<n and grid[nr][nc]!="#":
                    
                    new_keys=keys
                    if grid[nr][nc] in "abcdef":
                        new_keys = ''.join(sorted(set(keys + grid[nr][nc])))
                    
                    # If it's a door, check if we have the corresponding key
                    if grid[nr][nc] in "ABCDEF" and grid[nr][nc].lower() not in new_keys:
                        continue  # If we don't have the key, skip this move
                    
                    # If the new state hasn't been visited, process it
                    if (nr, nc, new_keys) not in visited:
                        visited.add((nr, nc, new_keys))
                        dist[(nr, nc, new_keys)] = dist[(r, c, keys)] + 1
                        q.append((nr, nc, new_keys))
                      
        
   
        return -1


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