<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>

from collections import deque 
DIRECTIONS = [(0, 1), (1, 0), (-1, 0), (0, -1)]
class Solution:
#     def shortestPathAllKeys(self, grid: List[str]) -> int:
# #         n, m = len(grid), len(grid[0])
# #         numOfKeys = 0 
# #         moves = set()
# #         for i in range(n):
# #             for j in range(m):
# #                 if grid[i][j] == '@':
# #                     starti = i 
# #                     startj = j
# #                 elif grid[i][j] in 'abcdef':
# #                     numOfKeys += 1
# #         queue = deque()
# #         queue.append([starti, startj, 0, '.@abcdef', 0])
# #         while queue:
# #             i, j, steps, keys, collectedKeys = queue.popleft()
# #             if grid[i][j] in 'abcdef' and grid[i][j].upper() not in keys:
# #                 keys += grid[i][j].upper()
# #                 collectedKeys += 1 
# #             if collectedKeys == numOfKeys:
# #                 return steps 
            
# #             for dx, dy in DIRECTIONS:
# #                 nx, ny = i+dx, j+dy 
# #                 if 0<= nx < n and 0<=ny< m and grid[nx][ny] in keys:
# #                     if (nx, ny, keys) not in moves:
# #                         moves.add((nx, ny, keys))
# #                         queue.append([nx, ny, steps+1, keys, collectedKeys])
                        
# #         return -1 

    def shortestPathAllKeys(self, grid: List[str]) -> int:
        n, m = len(grid), len(grid[0])
        numOfKeys = 0 
        for i in range(n):
            for j in range(m):
                if grid[i][j] == '@':
                    startx, starty = i, j 
                elif grid[i][j] in 'abcdef':
                    numOfKeys += 1
        minSteps = 0 
        visited = set()
        queue = deque([(startx, starty, minSteps,
                        '@.abcdef', 0)])
        while queue:
            cx, cy, minSteps, cur_keys, collectedKeys = \
            queue.popleft()
            if grid[cx][cy] in cur_keys and \
                grid[cx][cy].upper() not in cur_keys:
                cur_keys += grid[cx][cy].upper()
                collectedKeys += 1
            if numOfKeys == collectedKeys:
                return minSteps 
            for dx, dy in DIRECTIONS:
                nx, ny = cx+dx, cy+dy 
                if self.is_valid(nx, ny, grid) and \
                    grid[nx][ny] in cur_keys:
                    if (nx, ny, cur_keys) not in visited:
                        visited.add((nx, ny, cur_keys))
                        queue.append((nx, ny, minSteps+1,
                                        cur_keys, collectedKeys))
        return -1

    def is_valid(self, x, y, grid):
        n, m = len(grid), len(grid[0])
        if 0<=x<n and 0<=y<m:
            return True 
        return False 

    # def shortestPathAllKeys(self, grid: List[str]) -> int:
    #     n, m = len(grid), len(grid[0])
    #     numOfKeys = 0
    #     for i in range(n):
    #         for j in range(m):
    #             if grid[i][j] == '@':
    #                 startx, starty = i, j
    #             elif grid[i][j] in 'abcdef':
    #                 numOfKeys += 1
    #     minSteps = 0 
    #     visited = set()
    #     queue = deque([(startx, starty, minSteps,
    #                     '@.abcdef',0)])
    #     while queue:
    #         for _ in range(len(queue)):
    #             cx, cy, minSteps, cur_keys, collectedKeys = \
    #             queue.popleft()
    #             if grid[cx][cy] in cur_keys and \
    #                 grid[cx][cy].upper() not in cur_keys:
    #                 cur_keys += grid[cx][cy].upper()
    #                 collectedKeys += 1
    #             if numOfKeys == collectedKeys:
    #                 return minSteps 
    #             for dx, dy in DIRECTIONS:
    #                 nx, ny = cx+dx, cy+dy 
    #                 if self.is_valid(nx, ny, grid) and \
    #                     grid[nx][ny] in cur_keys:
    #                     if (nx, ny, cur_keys) not in visited:
    #                         visited.add((nx, ny, cur_keys))
    #                         queue.append((nx, ny, minSteps+1,
    #                                       cur_keys, collectedKeys))
    #         minSteps += 1
    #     return -1 
    
    # def is_valid(self, x, y, grid):
    #     n, m = len(grid), len(grid[0])
    #     if 0<=x<n and 0<=y<m:
    #         return True 
    #     return False 
        
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