<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:
        m, n = len(grid), len(grid[0])
        dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))

        in_bounds = lambda i, j: 0 <= i < m and 0 <= j < n
        nbrs = lambda i, j: ((i + di, j + dj) for di, dj in dirs)

        starfilter = lambda f, xs: filter(lambda args: f(*args), xs)
        starfilterfalse = lambda f, xs: filterfalse(lambda args: f(*args), xs)

        is_x = lambda val, x: val in x
        is_empty = partial(is_x, x='.')
        is_wall = partial(is_x, x='#')
        is_start = partial(is_x, x='@')
        is_key = partial(is_x, x='abcdef')
        is_lock = partial(is_x, x='ABCDEF')

        matrix2d_get = lambda i, j, mtx: mtx[i][j]
        get = partial(matrix2d_get, mtx=grid)

        k = sum(map(is_key, chain.from_iterable(grid))) # Total number of keys in grid.
        start = next(filter(lambda ij: is_start(get(*ij)), product(range(m), range(n))))
        
        frontier = deque([(start, 0, frozenset())])
        seen = {(indices, keys) for indices, d, keys in frontier}

        while frontier:
            (i, j), d, keys = frontier.popleft()

            nbrs_ = starfilter(in_bounds, nbrs(i, j)) # Keep only inbound nbrs.
            nbrs_ = filter(lambda nbr: (nbr, keys) not in seen, nbrs_) # Keep only nbr not seen with same keys.

            nbr_and_vals = map(lambda nbr: (nbr, get(*nbr)), nbrs_) # Pair up nbr and containing value.
            nbr_and_vals = starfilterfalse(lambda _, val: is_wall(val), nbr_and_vals) # Remove nbr with WALL as value.
            nbr_and_vals = starfilterfalse(lambda _, val: is_lock(val) and val.lower() not in keys, nbr_and_vals) # Remove nbr with LOCK to which we don't have key yet, as value. 

            for nbr, val in nbr_and_vals:
                new_d, new_keys = d + 1, (keys | {val} if is_key(val) else keys)
                if len(new_keys) == k: return new_d

                frontier.append((nbr, new_d, new_keys))
                seen.add((nbr, new_keys))
        
        return -1

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