<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>

def chooseN(lst: tuple, n: int) -> Generator[tuple, None, None]:
    l_len = len(lst)
    
    def recurr(i: int=0, m: int=n) -> Generator[tuple, None, None]:
        if m == 0:
            yield ()
            return
        if l_len - i < m: return
        if l_len - i == m:
            yield lst[i:]
            return
        for ans in recurr(i + 1, m - 1):
            yield (lst[i], *ans)
        for ans in recurr(i + 1, m):
            yield ans
        return
    for ans in recurr():
        yield ans
    return

import heapq
from sortedcontainers import SortedList

class Solution:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        if len(grid) == 0 or len(grid[0]) == 0: return 0
        shape = (len(grid), len(grid[0]))
        
        key_pos = {}
        for i, row in enumerate(grid):
            for j, l in enumerate(row):
                if l == "@": pos = (i, j)
                elif l.islower():
                    key_pos[l] = (i, j)
        keys = list(key_pos.keys())
        keys_dict = {k: i for i, k in enumerate(keys)}
        n_keys = len(keys)
        if n_keys == 0: return 0
        
        adjacency = {(x, y) for x in range(-1, 2) for y in range(-1, 2) if abs(x + y) == 1}
        def adjacent(pos: Tuple[int], k_list: List[int]) -> Generator[Tuple[int], None, None]:
            for d in adjacency:
                pos2 = tuple(x + y for x, y in zip(pos, d))
                for p, s in zip(pos2, shape):
                    if not 0 <= p < s: break
                else:
                    l = grid[pos2[0]][pos2[1]]
                    if l == "#" or l.isupper() and keys_dict[l.lower()] in k_list:
                        continue
                    yield pos2
            return
    
        def distance(pos1: Tuple[int], pos2: Tuple[int]) -> int:
            return sum(abs(x - y) for x, y in zip(pos1, pos2))
        
        key_dists = [{(i,): 0} for i in range(n_keys)]
        for i in range(n_keys):
            p1 = key_pos[keys[i]]
            for j in range(i + 1, n_keys):
                p2 = key_pos[keys[j]]
                dist = distance(p1, p2)
                key_dists[i][(i, j)] = dist
                key_dists[j][(i, j)] = dist
        
        for length in range(3, n_keys + 1):
            for k_lst in chooseN(tuple(range(n_keys)), length):
                for ind, i in enumerate(k_lst):
                    k_lst2 = (*k_lst[:ind], *k_lst[ind + 1:])
                    ans = float("inf")
                    for j in k_lst2:
                        ans = min(ans, key_dists[i][tuple(sorted((i, j)))] + key_dists[j][k_lst2])
                    key_dists[i][k_lst] = ans

        def heuristic(pos: Tuple[int], k_list: List[int]) -> int:
            res = float("inf")
            k_list = tuple(k_list)
            for i in k_list:
                res = min(res, distance(pos, key_pos[keys[i]]) + key_dists[i][k_list])
            return res
        
        k_list = SortedList(range(n_keys))
        heap = [(heuristic(pos, k_list), 0, pos, k_list)]
        seen = set((pos, tuple(k_list)))
        while heap:
            neg_d, pos, k_list = heapq.heappop(heap)[1:]
            tup = (pos, tuple(k_list))
            if tup in seen: continue
            neg_d -= 1
            seen.add(tup)
            for pos2 in adjacent(pos, k_list):
                l = grid[pos2[0]][pos2[1]]
                if l in keys_dict.keys() and keys_dict[l] in k_list:
                    if len(k_list) == 1: return -neg_d
                    k_list2 = SortedList(k_list)
                    k_list2.remove(keys_dict[l])
                else: k_list2 = k_list
                heapq.heappush(heap, (heuristic(pos2, k_list2) - neg_d, neg_d, pos2, k_list2))
        return -1
17548