<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 n,m,keys;
    int dx[4] = {1,-1,0,0};
    int dy[4] ={0,0,1,-1};
    int shortestPathAllKeys(vector<string>& g) {
        queue<pair<pair<int,int>,int>>q;
        n=g.size();
        m=g[0].size();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
            if(g[i][j]=='@')q.push({{i,j},0});
            if(g[i][j]>='a' && g[i][j]<='z')keys++;
        }
        }
        int check = (1<<keys)-1;
        map<pair<pair<int,int>,int>,int>p;
        int ans=0;
        while(q.size()){
            int k=q.size();
            for(int i=0;i<k;i++){
                auto j = q.front();q.pop();
                int x = j.first.first;
                int y = j.first.second;
                int mask = j.second;
                if(mask==check)return ans;
                for(int h=0;h<4;h++){
                    int nx = x + dx[h];
                    int ny = y + dy[h];
                    if(nx>=0 && ny>=0 && ny<m && nx<n && g[nx][ny]!='#'){
                        if(g[nx][ny]>='a' && g[nx][ny]<='z'){
                            int n_mask = mask | (1<<(g[nx][ny]-'a'));
                            if(p.find({{nx,ny},n_mask})==p.end()){
                                p[{{nx,ny},n_mask}]=n_mask;
                                q.push({{nx,ny},n_mask});
                            }
                        }
                        else if(g[nx][ny]>='A' && g[nx][ny]<='Z'){
                            int c = g[nx][ny]-'A';
                            if((mask & (1<<c)) == 0)continue;
                            int n_mask = mask;
                            if(p.find({{nx,ny},n_mask})==p.end()){
                               p[{{nx,ny},n_mask}]=n_mask;
                                q.push({{nx,ny},n_mask});
                            }
                        }
                        else{
                            int n_mask = mask;
                            if(p.find({{nx,ny},n_mask})==p.end()){
                                p[{{nx,ny},n_mask}]=n_mask;
                                q.push({{nx,ny},n_mask});
                            }
                        }
                    }
                }
            }
            ans++;
        }
        return -1;
    }
};
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