<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:
    def shortestPathAllKeys(self, grid: List[str]) -> int:
        map = {}
        print(grid)
        m = len(grid)
        n = len(grid[0])
        print(m, n)
        u = 0
        v = 0
        key = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '@':
                    u = i
                    v = j
                if grid[i][j] not in ['@', '.', '#'] and grid[i][j] == grid[i][j].lower():
                    key += 1
        print(u, v, key)
        q = []
        front = 0
        rear = 0
        map[''] = {}
        map[''][u] = {}
        map[''][u][v] = True
        q.append((u, v, '', 0, 0))
        dx = [0, 1, -1, 0]
        dy = [1, 0, 0, -1]
        while front <= rear:
            s = q[front]
            print(s)
            front += 1
            for k in range(4):
                x = s[0] + dx[k]
                y = s[1] + dy[k]
                if 0 <= x and x < m and 0 <= y and y < n:
                    if grid[x][y] != '#':
                        if grid[x][y] in ['.', '@']:
                            state = s[2]
                            if map.get(state, None) is None:
                                map[state] = {}
                            # end if
                            if map[state].get(x, None) is None:
                                map[state][x] = {}
                            # end if
                            if map[state][x].get(y, None) is None:
                                rear += 1
                                print('th1', grid[x][y], x, y)
                                q.append((x, y, state, s[3] + 1, s[4]))      
                                map[state][x][y] = True
                            # end if

                            
                        elif grid[x][y] != grid[x][y].upper(): #lower                            
                            state = ''.join(sorted(list(set(s[2] + grid[x][y]))))
                            if map.get(state, None) is None:
                                map[state] = {}
                            # end if
                            if map[state].get(x, None) is None:
                                map[state][x] = {}
                            # end if
                            if map[state][x].get(y, None) is None:
                                print('lower', grid[x][y], x, y)
                                rear += 1
                                q.append((x, y, state, s[3] + 1, s[4] + 1 if grid[x][y] not in s[2] else s[4] ))  
                                if ((s[4] + 1 if grid[x][y] not in s[2] else s[4]) == key):
                                    return s[3] + 1
                                map[state][x][y] = True
                            # end if
                        else: # upper
                            state = ''.join(sorted(list(set(s[2] + grid[x][y]))))
                            if map.get(state, None) is None:
                                map[state] = {}
                            # end if
                            if map[state].get(x, None) is None:
                                map[state][x] = {}
                            # end if
                            if map[state][x].get(y, None) is None and grid[x][y].lower() in s[2]:
                                print('upper', grid[x][y], x, y)
                                rear += 1
                                q.append((x, y, state, s[3] + 1, s[4]))      
                                map[state][x][y] = True
                            # end if
                        # end if
        return -1


        
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