<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,n = len(grid),len(grid[0])
        S = set(['A','B','C','D','E','F'])
        s = set(['a','b','c','d','e','f'])
        Lock = {}
        for i in range(m):
            for j in range(n):
                if grid[i][j] in s:
                    Lock[grid[i][j]] = len(Lock)
        dx,dy = [0,1,0,-1],[1,0,-1,0]
        Dis = {}
        visited = [[0]*n for i in range(m)]
        def bfs(interest):
            result = {}
            q = collections.deque()
            q.append(interest)
            d = 0
            while q:
                d+=1
                l = len(q)
                for i in range(l):
                    x,y = q.popleft()
                    for j in range(4):
                        nx,ny = x+dx[j],y+dy[j]
                        if 0<=nx<m and 0<=ny<n and (nx,ny) not in V:
                            V.add((nx,ny))
                            if grid[nx][ny] in s or grid[nx][ny] in S or grid[nx][ny]=='@':
                                result[grid[nx][ny]] = d
                            if grid[nx][ny] == '.':
                                q.append((nx,ny))
                Dis[grid[interest[0]][interest[1]]] = result
        for a in range(m):
            for b in range(n):
                if grid[a][b] in s or grid[a][b] in S or grid[a][b]=='@':
                    V =set()
                    V.add((a,b))
                    bfs((a,b))
        heap = [(0,0,'@')]
        mem = {(0,'@'):0}
        # print(Dis)
        # print(bin((1<<(len(Lock)))-1))
        while heap:
            top = heapq.heappop(heap)
            dis,state,at = top
            if state == (1<<(len(Lock)))-1:
                return dis
            for nei in Dis[at]:
                newdis = dis+Dis[at][nei]
                if nei in s:
                    news = state|(1<<Lock[nei])
                    if (news,nei) not in mem or mem[(news,nei)] > newdis:
                        # print('news',bin(news),'nei',nei,'newdis',newdis)
                        mem[(news,nei)] = newdis
                        heapq.heappush(heap,(newdis,news,nei))
                if nei in S:
                    if state&(1<<Lock[nei.lower()]) != 0 and ((state,nei) not in mem or mem[(state,nei)] > newdis):
                        # print('s',bin(state),'nei',nei,'newdis',newdis)
                        mem[(state,nei)] = newdis
                        heapq.heappush(heap,(newdis,state,nei))
                if nei == '@':
                    if (state,nei) not in mem or mem[(state,nei)] > newdis:
                        # print('at',bin(state),'nei',nei,'newdis',newdis)
                        mem[(state,nei)] = newdis
                        heapq.heappush(heap,(newdis,state,nei))
        return -1

                        
        

                        
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