<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:
        dr = [0, 0, -1, 1]
        dc = [1, -1, 0, 0]
        rows = len(grid)
        cols = len(grid[0])
        q = deque()
        start_r = start_c = None
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == '@':
                    start_r, start_c = r, c
                    q.append((r, c, set()))  
                    break
            if start_r is not None:
                break
        keys = {'a', 'b', 'c', 'd', 'e', 'f'}
        locks = {'A', 'B', 'C', 'D', 'E', 'F'}
        total_keys = sum(1 for r in range(rows) for c in range(cols) if grid[r][c] in keys)
        visited = set()
        visited.add((start_r, start_c, frozenset()))  
        steps = 0
        while q:
            for _ in range(len(q)):
                r, c, have_keys = q.popleft()
                if len(have_keys) == total_keys:
                    return steps
                for i in range(4):
                    rr = r + dr[i]
                    cc = c + dc[i]
                    if 0 <= rr < rows and 0 <= cc < cols:
                        cell = grid[rr][cc]
                        if cell == '#':
                            continue
                        new_keys = set(have_keys)
                        if cell in keys:
                            new_keys.add(cell)
                        if cell in locks:
                            if cell.lower() not in have_keys:
                                continue
                        state = (rr, cc, tuple(sorted(new_keys)))  
                        if state not in visited:
                            visited.add(state)
                            q.append((rr, cc, new_keys))
            steps += 1
        
        return -1

                        

                    

                        


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