<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 shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size(), n = m?grid[0].size():0;
        if(!m || !n)
            return -1;
        int target = 0;
        vector<int>start;
        for(int i = 0;i<m;i++){
            for(int j = 0;j<n;j++){
                if(isKey(grid[i][j]))
                    target |= 1<<(grid[i][j]-'a');
                else if(grid[i][j] == '@')
                    start = {i, j};
            }
        }
        if(!target)
            return 0;
        vector<int>dir{0, 1, 0, -1, 0};
        vector<vector<vector<int>>>visited(m, vector<vector<int>>(n, vector<int>(1<<6, 0)));
        queue<vector<int>>to_visit;
        to_visit.push({start[0], start[1], 0});
        visited[start[0]][start[1]][0] = 0;
        int lvl = 0;
        while(!to_visit.empty()){
            auto s = to_visit.size();
            while(s--){
                auto cur = to_visit.front();
                to_visit.pop();
                int x = cur[0], y = cur[1];
                if(cur[2] == target)
                    return lvl;
                for(int d = 0;d<dir.size()-1;d++){
                    int nx = x+dir[d], ny = y+dir[d+1], key = cur[2];
                    if(nx<0 || nx>= m || ny<0 || ny>=n || grid[nx][ny] == '#')
                        continue;
                    if(isLock(grid[nx][ny]) && !(key&(1<<(grid[nx][ny]-'A'))))
                        continue;
                    if(isKey(grid[nx][ny]))
                        key |= 1<<(grid[nx][ny]-'a');
                    if(!visited[nx][ny][key]){
                        visited[nx][ny][key] = 1;
                        to_visit.push({nx, ny, key});
                    }
                }
            }
            lvl++;
        }
        return -1;
    }
    
    bool isKey(char c){
        return c>='a' && c<='f';
    }
    bool isLock(char c){
        return c>='A' && c<='F';
    }
};
33000