<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:
        m, n = len(grid), len(grid[0])
        queue = collections.deque()
        
        # seen['key'] is only for BFS with key state equals 'keys'
        seen = collections.defaultdict(set)
        
        key_set, lock_set = set(), set()
        all_keys = 0
        start_r, start_c = -1, -1
        for i in range(m):
            for j in range(n):
                cell = grid[i][j]
                if cell in 'abcdef':
                    all_keys += (1 << (ord(cell) - ord('a')))
                    key_set.add(cell)
                if cell in 'ABCDEF':
                    lock_set.add(cell)
                if cell == "@":
                    start_r, start_c = i, j
        
        # [row, column, key_state, distance]
        queue.append((start_r, start_c, 0, 0))
        seen[0].add((start_r, start_c))
        
        while queue:
            cur_r, cur_c, keys, dist = queue.popleft()
            for dr, dc in ((0, 1), (1, 0), (-1, 0), (0, -1)):
                new_r, new_c = cur_r + dr, cur_c + dc
                
                # If this cell (new_r, new_c) is reachable.
                if 0 <= new_r < m and 0 <= new_c < n and grid[new_r][new_c] != '#':
                    cell = grid[new_r][new_c]
                    
                    # If it is a key we haven't picked up yet
                    if cell in key_set and not ((1 << (ord(cell) - ord('a'))) & keys):
                        new_keys = (keys | (1 << (ord(cell) - ord('a'))))

                        # If we collect all keys, return dist + 1. 
                        # Otherwise, just add this state to seen and queue.
                        if new_keys == all_keys:
                            return dist + 1
                        seen[new_keys].add((new_r, new_c))
                        queue.append((new_r, new_c, new_keys, dist + 1))
                      
                    # If it is a lock and we don't have its key, continue.
                    elif cell in lock_set and not (keys & (1 << (ord(cell) - ord('A')))):
                        continue
                        
                    # We can walk to this cell if we haven't been here before with the same key state.
                    elif (new_r, new_c) not in seen[keys]:
                        seen[keys].add((new_r, new_c))
                        queue.append((new_r, new_c, keys, dist + 1))
            
        return -1
19446