<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 = len(grid)
        n = len(grid[0])

        start_point = None
        all_keys = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j].islower():
                    shift = ord(grid[i][j]) - ord("a")
                    all_keys |= 1 << shift
                if grid[i][j] == "@":
                    start_point = (i, j)
        queue = deque()
        visited = set()
        queue.append((0, start_point, 0))

        while queue:
            keys_state, point, step = queue.popleft()
            if keys_state == all_keys:
                return step
            
            if (keys_state, point) in visited:
                continue
            visited.add((keys_state, point))

            for dx, dy in [(0,1), (0,-1), (1,0), (-1,0)]:
                x, y = point[0], point[1]
                next_x, next_y = x+dx, y+dy
                if (0 <= next_x < m) and (0 <= next_y <n):
                    char = grid[next_x][next_y]
                    if char in ('.', '@'):
                        queue.append((keys_state, (next_x, next_y), step+1))
                    elif char.islower():
                        shift = ord(char) - ord('a')
                        if (keys_state >> shift) & 1:
                            queue.append((keys_state, (next_x, next_y), step+1))
                        else:
                            new_keys_state = keys_state | 1 << shift
                            queue.append((new_keys_state, (next_x, next_y), step+1))
                    elif char.isupper():
                        shift = ord(grid[next_x][next_y]) - ord('A')
                        if (keys_state >> shift) & 1:
                            queue.append((keys_state, (next_x, next_y), step+1))
        return -1

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