<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:
    vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    int shortestPathAllKeys(vector<string>& grid) {
        queue<vector<int>> q; // {x, y, state}
        int h = grid.size(), w = grid[0].size();
        auto visited = vector<vector<unordered_set<int>>>(h, vector<unordered_set<int>>(w));
        int count = 0;
        for(int i = 0; i < h; ++i)
        {
            for(int j = 0; j < w; ++j)
            {
                if(grid[i][j] == '@')
                {
                    q.push({i, j, 0});
                    visited[i][j].insert(0);
                }    
                else if(grid[i][j] >= 'a' && grid[i][j] <= 'f') 
                    count++;           
            }
        }
        int finalState = 0;
        for(int i = 0; i < count; ++i)
            finalState |= (1 << i);
        int step = -1;
        while(!q.empty())
        {
            step++;
            int len = q.size();

            for(int i = 0; i < len; ++i)
            {
                int x = q.front()[0];
                int y = q.front()[1];
                int state = q.front()[2];
                q.pop();  

                for(auto d : dir)
                {
                    int xx = x + d.first;
                    int yy = y + d.second;
                    int newState = state;

                    if(xx < 0 || xx >= h || yy < 0 || yy >= w || grid[xx][yy] == '#')
                        continue;

                    if(grid[xx][yy] >= 'A' && grid[xx][yy] <= 'F' && (state >> (grid[xx][yy] - 'A') & 1) == 0)
                        continue;

                    if(grid[xx][yy] >= 'a' && grid[xx][yy] <= 'f')
                        newState |= 1 << (grid[xx][yy] - 'a');

                    if(visited[xx][yy].find(newState) != visited[xx][yy].end())
                        continue;

                    if(newState == finalState)
                        return step + 1;
                    
                    q.push({xx, yy, newState});
                    visited[xx][yy].insert(newState);
                }              
            }
        }

        return -1;
    }
};
34084