<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<vector<int> > dirs = {{0,1},{1,0},{0,-1},{-1,0}};
    int shortestPathAllKeys(vector<string>& grid) {
        int n = grid.size();
        int m = grid[0].size();

        int key_mask = 0;
        int start = 0;
        int end = 0;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j] == '@'){
                    start = i;
                    end = j;
                }
                if(grid[i][j] >= 'a' && grid[i][j] <= 'z'){
                    key_mask = key_mask | (1 << (grid[i][j] - 'a'));
                }
            }
        }

        bool visited[n][m][key_mask + 1];
        memset(visited, false, sizeof(visited));
        queue<vector<int>> q;
        q.push({start, end, 0, 0});
        visited[start][end][0] = true;

        while(!q.empty()){
            auto v = q.front();q.pop();
            int x = v[0], y = v[1], curr_mask = v[2], step = v[3];
            if(curr_mask == key_mask) return step;
            for(auto dir : dirs){
                int xx = x + dir[0];
                int yy = y + dir[1];
                if(xx >= 0 && xx < n && yy >= 0 && yy < m && grid[xx][yy] != '#'){
                    char ch = grid[xx][yy];
                    if(!visited[xx][yy][curr_mask] && ch >= 'A' && ch <= 'Z'){
                        bool curr_lock_key = curr_mask >> (ch - 'A') & 1;
                        cout<<"Current State: "<<xx<<yy<<" "<<curr_lock_key<<endl;
                        if(curr_lock_key){
                            q.push({xx,yy,curr_mask,step+1});
                            visited[xx][yy][curr_mask] = true;
                        }
                    }else if(ch >= 'a' && ch <= 'z'){
                        int new_mask = curr_mask | (1 << (ch - 'a'));
                        cout<<"New Mask: "<<xx<<yy<<" "<<new_mask<<endl;
                        if(!visited[xx][yy][new_mask]){
                            q.push({xx,yy,new_mask,step+1});
                            visited[xx][yy][new_mask] = true;
                        }
                    }else{
                        if(!visited[xx][yy][curr_mask]){
                            q.push({xx,yy,curr_mask,step+1});
                            visited[xx][yy][curr_mask] = true;
                        }
                    }
                }
            }
        }

        return -1;
    }
};
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