<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 n = grid.size();
        int m = grid[0].size();
        int keys = 0;
        queue<vector<int>> q; //{i,j,mask}
        for(int i = 0;i<n;i++)
        {
            for(int j =0;j<m;j++)
            {
                if(islower(grid[i][j]))keys++;
                else if(grid[i][j]=='@')
                {
                    q.push({i,j,0});
                }
            }
        } 
        int dx[4] = {-1, 1, 0, 0};
        int dy[4] = {0, 0, -1, 1};
        set<vector<int>>vis;
        int ans =0;
        while(!q.empty())
        {
            int sz = q.size();
            for(int i = 0;i<sz;i++)
            {
                vector<int> temp = q.front();
                q.pop();
                int x = temp[0];
                int y = temp[1];
                int mask = temp[2];
                if(__builtin_popcount(mask) == keys) return ans;
                for(int k = 0;k<4;k++)
                {
                    int nrow = x + dx[k];
                    int ncol = y + dy[k];
                    if(nrow < 0 || nrow >=n || ncol < 0 || ncol >=m || grid[nrow][ncol]=='#') continue;
                    int curr_mask = mask;
                    char ch = grid[nrow][ncol];
                    if(ch >= 'a' && ch <= 'z' )
                    {
                        curr_mask |= (1<< (ch - 'a'));
                    }
                    if(vis.find({nrow, ncol, curr_mask})!= vis.end()) continue;
                    if(ch >= 'A' && ch <= 'Z')
                    {
                        if((mask & (1<<(ch-'A')))== 0) continue;
                    }
                    q.push({nrow, ncol, curr_mask});
                    vis.insert({{nrow, ncol, curr_mask}});
                }
            }
            ans++;
        }
        return -1;
    }
};
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