<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:
bool dist[31][31][1<<6];
    int shortestPathAllKeys(vector<string>& grid) {
        int n=grid.size(),m=grid[0].size(),cnt=0,ind=0,ind2=0;
        unordered_map<char,int>m2;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]>='a' && grid[i][j]<='z'){m2[grid[i][j]]=cnt++;}
                if(grid[i][j]=='@'){ind=i;ind2=j;}
            }
        }
        int res=0;
        int mask=1<<cnt;
        int found=mask-1;
        queue<vector<int>>q;
        q.push({ind,ind2,0});
        while(!q.empty()){
            int n2=q.size();
            for(int i=0;i<n2;i++){
                auto p=q.front();
                q.pop();
                int a=p[0];
                int b=p[1];
                int mask=p[2];
                
                if (a < 0 || a>= n || b < 0 || b >= m)
                    continue;
                if (dist[a][b][mask])
                    continue;
                dist[a][b][mask]=1;
    
                if (grid[a][b] == '#')
                    continue;

                if (isupper(grid[a][b])) {
                    if ((mask & (1 << m2[tolower(grid[a][b])])) == 0)
                        continue;
                }
                if(grid[a][b]>='a' && grid[a][b]<='z'){
                    mask=mask|(1<<m2[grid[a][b]]);
                }
                if(mask==found){
                    return res;
                }
                q.push({ a + 1, b, mask });
                q.push({ a - 1, b, mask });
                q.push({ a, b + 1, mask });
                q.push({ a, b - 1, mask });
            }
            res++;
        }
        return -1;
    }
};
79601