<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
from typing import List

class Solution:
    def __init__(self):
        # Directions for moving in the grid: right, left, down, up
        self.directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
    
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        m, n = len(grid), len(grid[0])
        
        # Queue stores tuples of (x, y, steps, current_key_status_bitmask)
        que = deque()
        key_count = 0
        
        # Initialize the starting position and count the total number of keys
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '@':
                    que.append((i, j, 0, 0))  # Start position
                elif 'a' <= grid[i][j] <= 'f':
                    key_count += 1
        
        # The final state when all keys are collected
        final_key_status_bitmask = (1 << key_count) - 1
        
        # Visited array to track states: dimensions are (rows, cols, key_status)
        visited = [[[0] * (final_key_status_bitmask + 1) for _ in range(n)] for _ in range(m)]
        
        # BFS loop
        while que:
            x, y, steps, current_key_status_bitmask = que.popleft()
            
            # If all keys are collected, return the number of steps
            if current_key_status_bitmask == final_key_status_bitmask:
                return steps
            
            # Explore all 4 possible directions
            for dx, dy in self.directions:
                new_x, new_y = x + dx, y + dy
                
                if 0 <= new_x < m and 0 <= new_y < n and grid[new_x][new_y] != '#':
                    ch = grid[new_x][new_y]
                    
                    # If it's a lock and we have the corresponding key
                    if 'A' <= ch <= 'F' and ((current_key_status_bitmask >> (ord(ch) - ord('A'))) & 1):
                        if not visited[new_x][new_y][current_key_status_bitmask]:
                            visited[new_x][new_y][current_key_status_bitmask] = 1
                            que.append((new_x, new_y, steps + 1, current_key_status_bitmask))
                    
                    # If it's a key
                    elif 'a' <= ch <= 'f':
                        new_key_status_bitmask = current_key_status_bitmask | (1 << (ord(ch) - ord('a')))
                        if not visited[new_x][new_y][new_key_status_bitmask]:
                            visited[new_x][new_y][new_key_status_bitmask] = 1
                            que.append((new_x, new_y, steps + 1, new_key_status_bitmask))
                    
                    # If it's an empty cell or the start position
                    elif ch == '.' or ch == '@':
                        if not visited[new_x][new_y][current_key_status_bitmask]:
                            visited[new_x][new_y][current_key_status_bitmask] = 1
                            que.append((new_x, new_y, steps + 1, current_key_status_bitmask))
        
        # If no solution found, return -1
        return -1

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