<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:
        cache=set()
      
        k=0
        for a in range(0, len(grid)):
            for b in range(0, len(grid[0])):
                if grid[a][b] == '@':
                    pos = (a, b)
                elif ord('a') <= ord(grid[a][b]) <=ord('f'):
                    k+=1

        pos = pos+(frozenset(),)
        print(pos)
        q = deque()
        q.append(pos)
        cache.add(pos)
        count = 0
        while(q):
            print(q)
            size = len(q)
            count+=1
            for i in range(size):
                r, c, keys = q.popleft()
                dir = [[1,0], [0,1], [-1,0], [0,-1]]
                
                for a,b in dir:
                    nr = r+a
                    nc = c+b

                    if (0<=nr<len(grid) and 0<=nc<len(grid[0]) and (nr, nc, keys) not in cache and grid[nr][nc] != '#'):
                        ch = grid[nr][nc]

                        if ord('a') <= ord(ch) <=ord('f'):
                            new_keys = keys | frozenset([ch])
                            if len(new_keys) == k:
                                return count
                            
                            
                            cache.add((nr,nc, new_keys))
                            q.append((nr, nc, new_keys))

                        elif ord('A')<=ord(ch)<=ord('F'):
                            
                            if chr(ord(ch)+32) in keys:
                                q.append((nr, nc, keys))
                                cache.add((nr,nc, keys))

                        else:
                            q.append((nr, nc, keys))
                            cache.add((nr,nc, keys))
       
        return -1




        
        
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