<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:
        grid = list(map(list, grid))
        # print("original")
        # print(*grid, sep="\n")

        def remove_lock(ch: str) -> Tuple[int]:
            nonlocal grid

            for i in range(m):
                for j in range(n):
                    if grid[i][j] == ch:
                        # print("removed lock", ch)
                        grid[i][j] = "."
                        return (i, j)
            return (-1, -1)

        def is_key(ch: str) -> bool:
            return ord(ch) >= ord('a') and ord(ch) <= ord('z')

        def is_lock(ch: str) -> bool:
            return ord(ch) >= ord('A') and ord(ch) <= ord('Z')

        m = len(grid)
        n = len(grid[0])
        keys = 0
        src = None

        for i, s in enumerate(grid):
            for j, ch in enumerate(s):
                if is_key(ch):
                    keys += 1
                if ch == '@':
                    src = (i, j)
                    grid[i][j] = '.'

        def bfs(i: int, j: int) -> int:
            nonlocal keys, grid
            if not keys: return 0

            vis = set()
            q = deque([((i, j), 0)])

            ret = -1

            while q:
                x, moves = q.popleft()
                a, b = x
                if (a, b) in vis or is_lock(grid[a][b]) or grid[a][b] == '#':
                    continue
                vis.add((a, b))

                if is_key(grid[a][b]):
                    key = grid[a][b]
                    u, v = remove_lock(key.upper())
                    grid[a][b] = "."
                    keys -= 1

                    # print(moves, (a, b))
                    # print(*grid, sep="\n")
                    output = bfs(a, b)
                    if output != -1:
                        if ret == -1:
                            ret = moves + output
                        ret = min(ret, moves + output)

                    keys += 1
                    grid[u][v] = key.upper()
                    grid[a][b] = key

                if a != 0: q.append(((a - 1, b), moves + 1))
                if b != 0: q.append(((a, b - 1), moves + 1))
                if a != m - 1: q.append(((a + 1, b), moves + 1))
                if b != n - 1: q.append(((a, b + 1), moves + 1))

            return ret

        return bfs(*src)

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