<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])
        all_keys = []
        start_point = [0, 0]
        for i in range(M):
            for j in range(N):
                if grid[i][j] >= 'a' and grid[i][j] <= 'z':
                    all_keys.append(grid[i][j])
                if grid[i][j] == '@':
                    start_point = [i, j]

        all_keys.sort()
        K = len(all_keys)
        all_keys_mask = (1 << len(all_keys)) - 1
        all_keys_lookup = {}
        for i in range(K):
            all_keys_lookup[all_keys[i]] = i

        first = (start_point[0] << 12) | (start_point[1] << 6)
        q = deque([first])
        visited = set([first])
        moves = 0
        nexts = [[0,1],[0,-1],[1,0],[-1,0]]
        while q:
            k = len(q)
            for i in range(k):
                cur = q.popleft()
                x, y, keys = cur >> 12, (cur >> 6) & 0b111111, cur & 0b111111
                if keys == all_keys_mask:
                    return moves

                for dx, dy in nexts:
                    nx, ny = x+dx, y+dy
                    if nx < 0 or nx >= M or ny < 0 or ny >= N:
                        continue
                    if grid[nx][ny] == '#':
                        continue
                    if grid[nx][ny] >= 'A' and grid[nx][ny] <= 'Z':
                        to_key = chr(ord('a') + ord(grid[nx][ny]) - ord('A'))
                        if keys & (1 << all_keys_lookup[to_key]) == 0:
                            continue
                    next_keys = keys
                    if grid[nx][ny] >= 'a' and grid[nx][ny] <= 'z':
                        next_keys |= (1 << all_keys_lookup[grid[nx][ny]])

                    next = (nx << 12) | (ny << 6) | next_keys
                    if next not in visited:
                        visited.add(next)
                        q.append(next)
            moves += 1

        return -1


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