<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(object):
    def shortestPathAllKeys(self, grid):
        R, C = len(grid), len(grid[0])

        # The points of interest
        location = {v: (r, c)
                    for r, row in enumerate(grid)
                    for c, v in enumerate(row)
                    if v not in '.#'}

        def neighbors(r, c):
            for cr, cc in ((r-1, c), (r, c-1), (r+1, c), (r, c+1)):
                if 0 <= cr < R and 0 <= cc < C:
                    yield cr, cc

        # The distance from source to each point of interest
        def bfs_from(source):
            r, c = location[source]
            seen = [[False] * C for _ in range(R)]
            seen[r][c] = True
            queue = collections.deque([(r, c, 0)])
            dist = {}
            while queue:
                r, c, d = queue.popleft()
                if source != grid[r][c] != '.':
                    dist[grid[r][c]] = d
                    continue # Stop walking from here if we reach a point of interest
                for cr, cc in neighbors(r, c):
                    if grid[cr][cc] != '#' and not seen[cr][cc]:
                        seen[cr][cc] = True
                        queue.append((cr, cc, d+1))
            return dist        

        dists = {place: bfs_from(place) for place in location}
        target_state = 2 ** sum(p.islower() for p in location) - 1

        #Dijkstra
        pq = [(0, '@', 0)]
        final_dist = collections.defaultdict(lambda: float('inf'))
        final_dist['@', 0] = 0
        while pq:
            d, place, state = heapq.heappop(pq)
            if final_dist[place, state] < d: continue
            if state == target_state: return d
            for destination, d2 in dists[place].items():
                state2 = state
                if destination.islower(): #key
                    state2 |= (1 << (ord(destination) - ord('a')))
                elif destination.isupper(): #lock
                    if not(state & (1 << (ord(destination) - ord('A')))): #no key
                        continue

                if d + d2 < final_dist[destination, state2]:
                    final_dist[destination, state2] = d + d2
                    heapq.heappush(pq, (d+d2, destination, state2))

        return -1
        
16686