<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:
        row_len, col_len = len(grid), len(grid[0])
        poi = {}
        for r, row in enumerate(grid):
            for c, v in enumerate((row)):
                if v not in ".#":
                    poi[v] = (r, c)

        def neighbor(r, c):
            for i, j in [(r - 1, c), (r, c - 1), (r + 1, c), (r, c + 1)]:
                if 0 <= i < row_len and 0 <= j < col_len:
                    yield i, j

        def bfs(source, target, keys):
            sr, sc = poi[source]
            tr, tc = poi[target]
            seen = [[False] * col_len for _ in range(row_len)]
            seen[sr][sc] = True
            queue = deque([(sr, sc, 0)])
            while queue:
                r, c, d = queue.popleft()
                if r == tr and c == tc:
                    return d
                for nr, nc in neighbor(r, c):
                    if not seen[nr][nc] and grid[nr][nc] != "#":
                        if grid[nr][nc].isupper() and grid[nr][nc].lower() not in keys:
                            continue
                        queue.append((nr, nc, d + 1))
                        seen[nr][nc] = True
            return float("inf")

        ans = float("inf")
        keys = "".join(chr(ord("a") + i) for i in range(len(poi) // 2))
        for cand in permutations(keys):
            result = 0
            for i, target in enumerate(cand):
                source = "@" if i == 0 else cand[i - 1]
                d = bfs(source, target, cand[:i])
                result += d
                if result >= ans:
                    break
            ans = min(result, ans)
        return ans if ans < float("inf") else -1

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