<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 di[4]={-1, 0, 1, 0};
    int dj[4]={0, 1, 0, -1};

    int shortestPathAllKeys(vector<string>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int keys = 0; // number of keys
        int startx, starty;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                char c = grid[i][j];
                if(c == '@')
                {
                    startx = i;
                    starty = j;
                }
                else if(c>='a' && c <= 'z')
                {
                    keys++;
                }
            }
        }
        // used i, j, keys
        vector<vector<vector<bool>>> visited(n, vector<vector<bool>>(m, vector<bool>(1<<keys)));
        visited[startx][starty][0]=true;
        queue<vector<int>> q;
        q.push({startx, starty, 0});
        int dist=0;
        while(!q.empty())
        {
            for(int size = q.size(); size>0; --size)
            {
                vector<int> cur = q.front();
                q.pop();
                int i = cur[0];
                int j = cur[1];
                int keyState = cur[2];
                if(keyState == (1<<keys)-1)
                {
                    return dist;
                }
                for(int k=0; k<4; k++)
                {
                    int i1 = i+di[k];
                    int j1 = j+dj[k];
                    if(i1>=0 && i1 <n && j1>=0 && j1<m )
                    {
                        char val = grid[i1][j1];
                        if(val=='#'|| (val>='A'&&val<='Z'&&!(keyState>>(val-'A')&1)))
                        {
                            continue;
                        }
                        int nextKeyState = keyState;
                        if(val>='a'&&val<='z')
                        {
                            nextKeyState |= 1<<(val-'a');
                        }
                        if(!visited[i1][j1][nextKeyState])
                        {
                            visited[i1][j1][nextKeyState] = true;
                            q.push({i1, j1,nextKeyState});
                        }
                    }
                }  
            }
            dist++;
        }
        return -1; 
    }
};
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