<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:

    class Path:
        def __init__ (self, curr_position, num_moves, acquired_keys):
            self.curr_position = curr_position
            self.num_moves = num_moves
            self.acquired_keys = acquired_keys
            
            # self.seen_squares = set ()

        def add_key (self, key):
            key_num = ord (key) - ord ('a')

            new_num = 1 << key_num

            self.acquired_keys |= new_num
            

    
    def shortestPathAllKeys (self, grid: List[str]) -> int:

        kDeltas = [(-1, 0), (1, 0), (0, -1), (0, 1)]

        kLockKeyMap = {chr (ord ('A') + i) : chr (ord ('a') + i) for i in range (6)}

        num_rows = len (grid)
        num_cols = len (grid[0])

        seen = [[set () for x in range (num_cols)] for y in range (num_rows)]
        
        
        num_keys = 0

        start = None

        all_keys = Solution.Path (0, 0, 0)
        
        for i, s in enumerate (grid):
            for j, c in enumerate (s):
                if c.islower ():
                    num_keys += 1
                    all_keys.add_key (c)

                elif c == '@':
                    start = (i, j)

        p1 = Solution.Path (start, 0, 0)

        # print ("all keys: ", all_keys.acquired_keys)

        # seen[start[0]][start[1]].
        
        q = deque ([p1])

        while q:

           

            top = q.pop ()
            
            

            for row_delta, col_delta in kDeltas:

                new_row = top.curr_position[0] + row_delta
                new_col = top.curr_position[1] + col_delta

                
                if (new_row < 0 or new_row >= num_rows or new_col < 0 or new_col >= num_cols or
                    grid[new_row][new_col] == '#'):

                    continue

                if top.acquired_keys in seen[new_row][new_col]:
                    continue

                

                

                c = grid[new_row][new_col]
                
                sp = Solution.Path ((new_row, new_col), top.num_moves + 1, top.acquired_keys)
                if c.isalpha ():
                    if c.isupper ():
                        key = kLockKeyMap[c]
                        key_num = 1 << (ord (key) - ord ('a'))
                        if (key_num & sp.acquired_keys) == 0:
                            continue

                    else:
                        sp.add_key (c)

               
                if all_keys.acquired_keys == sp.acquired_keys:
                    return sp.num_moves

            
                seen[new_row][new_col].add (sp.acquired_keys)

                
                
                q.appendleft (sp)
                
        
        return -1

                
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