<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 string
import heapq
from collections import deque
class Solution:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        self.num_of_keys = 0
        keys_coordinates = []
        starting_point = False
        n,m = len(grid), len(grid[0])
        moves = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        for i in range(n):
            for j in range(m):
                if grid[i][j] == '@':
                    starting_point = (i,j)
                elif grid[i][j] in ascii_lowercase:
                    self.num_of_keys+=1
                    keys_coordinates.append((i,j))
        
        print(keys_coordinates)
        def bfs(start_x, start_y):
            # Queue will store tuples of (x, y, distance, keys_collected)
            q = deque([(start_x, start_y, 0, set())])
            # Set to keep track of visited states: (x, y, keys_collected)
            seen = set([(start_x, start_y, frozenset())])
            
            while q:
                x, y, dist, keys_collected = q.popleft()
                
                for dx, dy in moves:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < n and 0 <= ny < m and grid[nx][ny] != '#':
                        new_keys = set(keys_collected)
                        
                        # If it's a key, collect it
                        if grid[nx][ny].islower():
                            new_keys.add(grid[nx][ny])
                        
                        # If all keys are collected, return the distance
                        if len(new_keys) == self.num_of_keys:
                            return dist + 1
                        
                        # If it's a door, check if we have the corresponding key
                        if grid[nx][ny].isupper() and grid[nx][ny].lower() not in new_keys:
                            continue  # Skip if we don't have the corresponding key
                        
                        # Make sure to only revisit states that are different
                        state = (nx, ny, frozenset(new_keys))
                        if state not in seen:
                            seen.add(state)
                            q.append((nx, ny, dist + 1, new_keys))
            
            return -1  # Impossible to collect all keys

        start_x, start_y = starting_point
        return bfs(start_x, start_y)


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