<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>


# Submitted by Samy Vilar <samy_vilar> on 09/05/2024

# O(n * m * 2**k) time w/ O(n * m * 2**k) additional-space
# through BFS; essentially/informally while applying BFS we continue
# exploring all never befored witnessed shortest paths with respect to 
# each and every possible subset of keys picked up; halting as soon as
# we picked up all keys;

# Since any cell is enqued/push at most 2**k possible times while for
# each of the possible 2**k subset keys we are keeping track of at most
# n * m possible cordinates would imply the aforementioned complexities;

# version 1.2 micro-optimization(s)/simplication
#            - flattened/sorrounded grid
#            - normalized keys w/ respect to distinct binary digit greater than 1
#            - let '@' and '.' have an id of 1, for simplication;

# version 1.2.1 used greater binary digit instead of 1 for '@' and '.'
#               cutting the additional-space in half; though at the cost
#               of some additional (implementation) complexity;

# version 1.2.2 checked none of the keys are completely walled off otherwise
#               we can safely yield -1;
#               rearranged conditional(s)
#               normalized both keys and locks distinguising the the by their
#               sign;

class Solution:
    def shortestPathAllKeys(self, cells: List[str], chained=chain.from_iterable) -> int:        
        cells = tuple(chain(chained(chained(zip(cells, repeat('#')))), 
                            repeat('#', deltas := len(cells[0]) + 1)))            
        seen = {root := cells.index('@')}
        pending = deque((root,))         
        deltas = 1, deltas, -1, -deltas       
        cnt = sum(map(bool, filter('a'.__le__, cells)))
        ids = {}
        targets = 1
        while True:            
            for at in map(pending.popleft().__add__, deltas):
                if (cell := cells[at]) != '#' and at not in seen:
                    if 'a' <= (cell := cells[at]):                        
                        ids[cell] = targets
                        ids[cell.upper()] = -targets
                        targets += targets
                        if cnt < targets.bit_length():
                            break
                    seen.add(at)
                    pending.appendleft(at)                    
            else:
                if not pending:
                    return -1
                continue
            break
        ids['#'] = 0
        ids['.'] = ids['@'] = upper = targets
        cells = tuple(map(ids.__getitem__, cells))      
        pending.clear()  
        pending.appendleft((root, targets))
        dests = tuple(starmap(set, repeat((), targets)))
        dests[0].add(root)
        pending.appendleft((root, targets))                
        targets += targets - 1        
        for moves in count(): 
            for at, keys in starmap(pending.popleft, repeat((), len(pending))):
                for at in map(at.__add__, deltas):                    
                    if (cell := cells[at]) < 0:                        
                        if -cell & keys and at not in (seen := dests[keys - upper]):
                            seen.add(at)
                            pending.append((at, keys))                    
                    elif cell:
                        if cell & (forward := keys) or (forward := forward + cell) != targets:
                            if at not in (seen := dests[forward - upper]):
                                seen.add(at)
                                pending.append((at, forward))
                        else:
                            return moves + 1 
            if not pending:
                return -1            

19101