<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>

import heapq

class Solution:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        n, m = len(grid), len(grid[0])
        pq = []

        # path_len, cur_coord, subset_of_keys
        start_coord = (0, 0)
        num_of_keys = 0
        for i in range(n):
            for j in range(m):
                if grid[i][j] == "@":
                    start_coord = (i, j)
                elif 0 <= ord(grid[i][j]) - ord('a') < 6:
                    num_of_keys = max(num_of_keys, ord(grid[i][j]) - ord('a') + 1)

        heapq.heappush(pq, (0, start_coord, 0))
        vis = [[[False for _ in range(1 << num_of_keys)] for _ in range(m)] for _ in range(n)]

        while pq:
            path_len, cur_coord, subset_of_keys = heapq.heappop(pq)
            cur_y, cur_x = cur_coord

            if subset_of_keys == (1 << num_of_keys) - 1:
                return path_len

            if vis[cur_y][cur_x][subset_of_keys]:
                continue

            vis[cur_y][cur_x][subset_of_keys] = True

            for y, x in [(cur_y-1, cur_x), (cur_y+1, cur_x), (cur_y, cur_x-1), (cur_y, cur_x+1)]:
                if not (0 <= y < n and 0 <= x < m) or grid[y][x] == '#':
                    continue
                elif (grid[y][x] == '.' or grid[y][x] == '@') and not vis[y][x][subset_of_keys]:
                    heapq.heappush(pq, (path_len + 1, (y, x), subset_of_keys))
                elif 0 <= ord(grid[y][x]) - ord('a') < 6:
                    # landed on key
                    key =  ord(grid[y][x]) - ord('a')
                    new_subset_of_keys = subset_of_keys | (1 << key)
                    if not vis[y][x][new_subset_of_keys]:
                        heapq.heappush(pq, (path_len + 1, (y, x), new_subset_of_keys))
                elif 0 <= ord(grid[y][x]) - ord('A') < 6:
                    # landed on key
                    key =  ord(grid[y][x]) - ord('A')
                    if subset_of_keys & (1 << key) and not vis[y][x][subset_of_keys]:
                        heapq.heappush(pq, (path_len + 1, (y, x), subset_of_keys))

        return -1
17376