<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) {
        // make mask as visited to, to track all the paths
        int n = grid.size(), m = grid[0].size();
        queue<pair<int,pair<int,int>>> q;
        int cnt = 0;
        int sx = -1, sy = -1;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(grid[i][j] == '@'){
                    sx = i, sy = j;
                }else if(islower(grid[i][j])) 
                    cnt++;
            }
        }
        int finalState = (1 << cnt) - 1;
        vector<vector<vector<int>>> vis(n, vector<vector<int>>(m, vector<int>(1 << cnt)));
        vis[sx][sy][0] = 1;
        q.push({0, {sx, sy}});
        int dir[] = {0, 1, 0, -1, 0};
        int step = 0;
        while(!q.empty()){
            int sz = q.size();
            for(int i=0; i<sz; i++){
                int x = q.front().second.first;
                int y = q.front().second.second;
                int key = q.front().first;
                q.pop();
                if(key == finalState) return step;
                for(int j=0; j<4; j++){
                    int nr = x + dir[j];
                    int nc = y + dir[j+1];
                    if(nr >= 0 && nr < n && nc >= 0 && nc < m && grid[nr][nc] != '#'){
                        if(islower(grid[nr][nc])){
                            int nkey = key | (1 << (grid[nr][nc] - 'a'));
                            if(!vis[nr][nc][nkey]){
                                vis[nr][nc][nkey] = 1;
                                q.push({nkey, {nr, nc}});
                            }
                        }else if(isupper(grid[nr][nc])){
                            if(key & (1 << (grid[nr][nc] - 'A')) && !vis[nr][nc][key]){
                                vis[nr][nc][key] = 1;
                                q.push({key, {nr, nc}});
                            }
                        }else{
                            if(!vis[nr][nc][key]){
                                vis[nr][nc][key] = 1;
                                q.push({key, {nr, nc}});
                            }
                        }
                    }
                }
            }
            step++;
        }
        return -1;
    }
};
15660