<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:
        # still feels like BFS, prob just consider same node visits with different keys found as different cases (unvisited)
        # queue items would have i, j for coords, an string of found keys, and the number of steps to get there
        queue = []
        keyCnt=0
        xDir = [1, -1, 0, 0]
        yDir = [0, 0, 1, -1]
        keys = ['a', 'b', 'c', 'd', 'e', 'f']
        locks = ['A', 'B', 'C', 'D', 'E', 'F']
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]=='@':
                    queue.append((i,j,'',0))
                elif grid[i][j] in keys:
                    keyCnt+=1
        visited = set()
        while len(queue) > 0:
            i,j,foundKeys,stepCnt = queue[0]
            queue.pop(0)
            tileVal=grid[i][j]
            if (tileVal in locks and tileVal.lower() not in foundKeys) or tileVal=='#':
                continue
            if tileVal in keys and tileVal not in foundKeys:
                foundKeys = foundKeys + tileVal
                if len(foundKeys) == keyCnt:
                    return stepCnt
            for dir in range(0, 4):
                newX = i + xDir[dir]
                newY = j + yDir[dir]
                if 0 <= newX < len(grid) and 0 <= newY < len(grid[0]) and (newX,newY,foundKeys) not in visited:
                    visited.add((newX,newY,foundKeys))
                    queue.append((newX,newY,foundKeys,stepCnt+1))
        return -1                                            
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