<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, namedtuple

State = namedtuple('State', ['row', 'col', 'keys'])

class Solution:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        total_keys = 0
        for row in range(len(grid)):
            for col in range(len(grid[0])):
                if grid[row][col] == '@':
                    start_state = State(row=row, col=col, keys=tuple([False] * 26))
                if 'a' <= grid[row][col] <= 'z':
                    total_keys += 1

        qu = deque([(start_state, 0)])
        visited = {start_state}
        dirs = [(-1, 0), (1, 0), (0, -1), (0, 1)]

        while qu:
            (row, col, keys), moves = qu.pop()

            next_keys = list(keys)
            if 'a' <= grid[row][col] <= 'z':
                next_keys[ord(grid[row][col]) - ord('a')] = True
            next_keys = tuple(next_keys)

            if sum(next_keys) == total_keys:
                return moves
                
            for dr, dc in dirs:
                nr, nc = row + dr, col + dc
                next_state = State(row=nr, col=nc, keys=next_keys)
                if (
                    0 <= nr < len(grid)
                    and 0 <= nc < len(grid[0])
                    and grid[nr][nc] != '#'
                    and next_state not in visited
                    and (not 'A' <= grid[nr][nc] <= 'Z' or next_keys[ord(grid[nr][nc]) - ord('A')])
                ):
                    visited.add(next_state)
                    qu.appendleft((next_state, moves + 1))


        return -1
25483