<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, n = len(grid), len(grid[0])
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        # Find the starting position and count the total number of keys
        start_x, start_y = 0, 0
        key_count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '@':
                    start_x, start_y = i, j
                elif grid[i][j].islower():
                    key_count += 1

        # BFS queue and visited set
        queue = deque([(start_x, start_y, 0, set())])  # (x, y, moves, keys)
        visited = set((start_x, start_y, frozenset()))

        while queue:
            x, y, moves, keys = queue.popleft()

            # If we have collected all keys, return the number of moves
            if len(keys) == key_count:
                return moves

            # Explore all four directions
            for dx, dy in directions:
                nx, ny = x + dx, y + dy

                # Check if the new position is within bounds
                if 0 <= nx < m and 0 <= ny < n:
                    cell = grid[nx][ny]

                    # If it's a wall, skip
                    if cell == '#':
                        continue

                    # If it's a lock and we don't have the key, skip
                    if cell.isupper() and cell.lower() not in keys:
                        continue

                    # If it's a key, add it to the set of keys
                    new_keys = keys.copy()
                    if cell.islower():
                        new_keys.add(cell)

                    # If the state has not been visited, add it to the queue and visited set
                    if (nx, ny, frozenset(new_keys)) not in visited:
                        visited.add((nx, ny, frozenset(new_keys)))
                        queue.append((nx, ny, moves + 1, new_keys))

        # If we couldn't collect all keys, return -1
        return -1

# Example usage:
solution = Solution()
grid = ["@.a..", "###.#", "b.A.B"]
print(solution.shortestPathAllKeys(grid))  # Output: 8
26518