<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 {
public:

    int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};

    int shortestPathAllKeys(vector<string>& grid) {
        
        int m = grid.size();
        int n = grid[0].size();
        queue<pair<pair<int, int>, int>> que;
        unordered_map<int,unsigned long> mapKey;
        int target = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if ('a' <= grid[i][j] && grid[i][j] <= 'f') {
                    target |= 1 << (grid[i][j] - 'a');
                    continue;
                }

                if (grid[i][j] == '@') {
                    que.push({{i, j}, 0});
                    mapKey[i*31+j] = 1;
                }
            }
        }


        int step = 0;
        while (!que.empty()) {
            int size = que.size();
            while(size--) {
                int i = que.front().first.first;
                int j = que.front().first.second;
                int key = que.front().second;
                que.pop();
                
                if ('a' <= grid[i][j] && grid[i][j] <= 'f') {
                    key |= 1 << (grid[i][j] - 'a');
                    if (key == target)
                        return step;

                    mapKey[i*31+j] |= 1l << key;
                }

                for (int d = 0; d < 4; d++) {
                    int newi = i + dir[d][0];
                    int newj = j + dir[d][1];
                    
                    if (!(0 <= newi && newi < m && 0 <= newj && newj < n))
                        continue;
                        
                    int val = grid[newi][newj];
                    if (val == '#')
                        continue;
                        
                    if ('A' <= val && val <= 'F' && (key & (1 << (val - 'A'))) == 0)
                        continue;
                        
                    if (mapKey[newi*31+newj] & (1l << key))
                        continue;
                        
                     mapKey[newi*31+newj] |= 1l << key;
                     que.push({{newi, newj}, key});
                }
            }
            step++;
        }
        return -1;
    }
};
14576