<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:
        # (row, col, current keys, # moves)
        visited = set()
        starting_row = None
        starting_col = None
        total_num_keys = 0
        for row in range(0, len(grid)):
            for col in range(0, len(grid[0])):
                if (grid[row][col] == "@"):
                    starting_row = row
                    starting_col = col
                elif (isKey(grid[row][col])):
                    total_num_keys += 1

        queue = [(starting_row, starting_col, tuple([0 for i in range(0, total_num_keys)]), 0)]
        count = 0
        while queue:
            new_queue = []
            for entry in queue:
                row, col, curr_keys, moves = entry
                if ((row, col, curr_keys) not in visited):
                    visited.add((row, col, curr_keys))
                    if (hasAllKeys(curr_keys)):
                        return moves
                    for add_row in range(-1, 2):
                        for add_col in range(-1, 2):
                            if (abs(add_row) != abs(add_col)):
                                if (checkInBounds(row+add_row, col+add_col, grid) and not isWall(grid[row + add_row][col + add_col])):
                                    next_keys = list(curr_keys)
                                    if (isKey(grid[row+add_row][col+add_col])):
                                        next_keys[ord(grid[row+add_row][col+add_col]) - ord("a")] = 1
                                    next_keys = tuple(next_keys)
                                    if (not isLock(grid[row + add_row][col + add_col]) or canLockOpen(grid[row + add_row][col + add_col], curr_keys)):
                                        if ((row+add_row, col + add_col, next_keys) not in visited):
                                            new_entry = (row + add_row, col + add_col, next_keys, moves + 1)
                                            new_queue.append(new_entry)
            queue = new_queue
            count += 1
            # if (count == 10):
            #     print("here")
            #     break
        return -1

def checkInBounds(new_row, new_col, grid):
    if (new_row < 0 or new_row >= len(grid) or new_col < 0 or new_col >= len(grid[0])):
        return False
    else:
        return True

def hasAllKeys(curr_keys):
    return all(curr_keys)

def isLock(element):
    if (element == "A" or element == "B" or element == "C" or element == "D" or element == "E" or element == "F"):
        return True
    else:
        return False

def canLockOpen(lockElement, curr_keys):
    return curr_keys[ord(lockElement) - ord("A")] == 1

def isKey(element):
    if (element == "a" or element == "b" or element == "c" or element == "d" or element == "e" or element == "f"):
        return True
    else:
        return False

def isWall(element):
    return element == "#"
    
        
24276