<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>

from collections import deque
class Solution:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        m = len(grid[0])
        n = len(grid)
        start = (0, 0, 0)
        keys = ['a', 'b', 'c', 'd', 'e', 'f']
        locks = ['A', 'B', 'C', 'D', 'E', 'F']
        num_keys = 0
        min_dist = None
        for i, s in enumerate(grid):
            for j, c in enumerate(s):
                if c == '@':
                    start = (i, j, 0)
                if c in keys:
                    num_keys += 1
        

        def bfs(start, key_chain):
            queue = deque()
            visited = set()
            queue.append(start)
            while queue:
                x, y, d = queue.popleft()
                if all(key_chain):
                    nonlocal min_dist
                    if min_dist == None:
                        min_dist = d
                    min_dist = min(min_dist, d)
                if x < 0 or y < 0 or x >= n or y >= m:
                    continue
                if (x, y) in visited:
                    continue
                visited.add((x, y))
                if grid[x][y] == '#':
                    continue
                if grid[x][y] in locks:
                    if not key_chain[locks.index(grid[x][y])]:
                        continue
                if grid[x][y] in keys and not key_chain[keys.index(grid[x][y])]:
                    new_chain = list(key_chain)
                    new_chain[keys.index(grid[x][y])] = True
                    bfs((x, y, d), new_chain)
                    
                queue.append((x + 1, y, d + 1))
                queue.append((x - 1, y, d + 1))
                queue.append((x, y + 1, d + 1))
                queue.append((x, y - 1, d + 1))

        bfs(start, [False] * num_keys)

        if min_dist:
            return min_dist
        else:
            return -1
17376