<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:
        directions = [(-1, 0), (0, -1), (1, 0), (0, 1)]
        queue = deque([])
        numKeys = 0
        startIndex = ()
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] == '@':
                    startIndex = (i, j)
                elif grid[i][j].isalpha() and grid[i][j].islower():
                    numKeys += 1
        keys = (0, 0, 0, 0, 0, 0)
        queue.append((startIndex[0], startIndex[1], 0, keys))

        indexes = {
            'a': 0,
            'b': 1,
            'c': 2,
            'd': 3,
            'e': 4,
            'f': 5
        }

        visited = set()
        while len(queue) > 0:
            row, col, pathCount, keysFound = queue.popleft()
            if not(0 <= row < len(grid) and 0 <= col < len(grid[row])):
                continue
            if (row, col, keysFound) in visited:
                continue
            elif grid[row][col] == "#":
                continue
            newKeysFound = list(keysFound)
            if grid[row][col].isalpha():
                if grid[row][col].islower():
                    # found a key
                    newKeysFound[indexes[grid[row][col]]] = 1
                elif keysFound[indexes[grid[row][col].lower()]] != 1:
                    # we don't have the key for this lock
                    continue
            visited.add((row, col, tuple(newKeysFound)))
            if newKeysFound.count(1) == numKeys:
                return pathCount
            for r, c in directions:
                queue.append((row + r, col + c, pathCount + 1, tuple(newKeysFound)))
            
        return -1
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