<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:
        lockkey = {"A":"a","B":"b","C":"c","D":"d","E":"e","F":"f"}
        memo = defaultdict(lambda: defaultdict(lambda: defaultdict(lambda: math.inf)))
        m,n = len(grid),len(grid[0])
        hp = []
        key = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == "@":
                    heapq.heappush(hp,(0,i,j,""))
                    memo[""][i][j] = 0
                elif grid[i][j].islower():
                    key += 1
        dirr = [(1,0),(-1,0),(0,1),(0,-1)]
        while hp:
            dist,x,y,keyset = heapq.heappop(hp)
            print("dist,x,y,keyset",dist,x,y,keyset)
            if len(keyset) == key:
                return dist
            for i in range(4):
                dx,dy = dirr[i]
                if not (0 <= x+dx < m and 0 <= y+dy < n): continue
                if grid[x+dx][y+dy] == "#": continue
                if grid[x+dx][y+dy] == "." or grid[x+dx][y+dy] == "@": 
                    if dist+1 >= memo[keyset][x+dx][y+dy]: continue
                    memo[keyset][x+dx][y+dy] = dist+1
                    heapq.heappush(hp,(dist+1,x+dx,y+dy,keyset))
                elif grid[x+dx][y+dy].isupper():
                    if lockkey[grid[x+dx][y+dy]] not in keyset: continue
                    if dist+1 >= memo[keyset][x+dx][y+dy]: continue
                    memo[keyset][x+dx][y+dy] = dist+1
                    heapq.heappush(hp,(dist+1,x+dx,y+dy,keyset))
                elif grid[x+dx][y+dy].islower():
                    if grid[x+dx][y+dy] in keyset:
                        if dist+1 >= memo[keyset][x+dx][y+dy]: continue
                        memo[keyset][x+dx][y+dy] = dist+1
                        heapq.heappush(hp,(dist+1,x+dx,y+dy,keyset))
                    else:
                        memo[keyset+grid[x+dx][y+dy]][x+dx][y+dy] = dist+1
                        heapq.heappush(hp,(dist+1,x+dx,y+dy,keyset+grid[x+dx][y+dy]))
        return -1
                    
                    




        
21861