<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:
    typedef pair<int, int> P; 
    vector<vector<int>> dir = { {1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    int bfs(pair<int, int>& src, vector<string>& grid, int countKeys, int m, int n)
    {
        queue<vector<int>> qu; // {i, j, distance, keys}
        unordered_map<string, bool> visited; 
        
        qu.push({src.first, src.second, 0, 0});
        visited["0_0_0"] = true; // x_y_key; 

        while(!qu.empty())
        {
            int x = qu.front()[0]; 
            int y = qu.front()[1];
            int dist = qu.front()[2];
            int key = qu.front()[3];
            // cout << x << " " << y << " " << dist <<" " << key << endl; 
            qu.pop(); 
            if(key == countKeys)
                return dist; 
            for(int i = 0;i < 4; i++)
            {
                int nx = x + dir[i][0]; 
                int ny = y + dir[i][1];
                string vis = to_string(nx) + "_" + to_string(ny);  
                if(nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] != '#')
                {
                    // if there is lock, check if corresponding key is present. 
                    if(grid[nx][ny] >= 'A' && grid[nx][ny] <= 'F')
                    {
                        int lockPos = grid[nx][ny] - 'A';
                        vis += "_" + to_string(key); 
                        if(((key >> lockPos) & 1) && !visited[vis])
                        {
                            qu.push({nx, ny, dist+1, key});
                            visited[vis] = true; 
                            // cout <<"neighbour 1 " << nx << " " << ny << " " << dist+1 <<" " << key << endl; 
                        }

                    }
                    // if there is key, then take that key. 
                    else if(grid[nx][ny] >= 'a' && grid[nx][ny] <= 'f')
                    {
                        int keyPos = grid[nx][ny] - 'a';
                        int mask = 1; 
                        mask <<= keyPos;
                        int temp = key | mask; 
                        vis += "_" + to_string(key); 
                        if(!visited[vis])
                        {
                            qu.push({nx, ny, dist+1, temp}); 
                            visited[vis] = true; 
                            //  cout <<"neighbour 2 " << nx << " " << ny << " " << dist+1 <<" " << key << endl; 
                        }

                    }
                    // if it is an empty cell 
                    else
                    {
                        vis += "_" + to_string(key); 
                        if(!visited[vis])
                        {
                            qu.push({nx, ny, dist+1, key});
                            visited[vis] = true; 
                            cout <<"neighbour 3 " << nx << " " << ny << " " << dist+1 <<" " << key << endl; 
                        }
                    }
                }
            }
        }
        return -1;  
    }
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size(); 
        int n =  grid[0].size(); 
        int countKeys = 0;
        pair<int, int> src;
        for(int i = 0;i < m; i++)
            for(int j = 0;j < n; j++)
                {
                    if(grid[i][j] == '@')
                    {
                        src = {i, j};
                    }
                    else if(grid[i][j] >= 'a' && grid[i][j] <= 'f')
                        countKeys ++; 
                }
        countKeys = pow(2,countKeys)-1; 
        int minMoves = bfs(src, grid, countKeys, m, n);
        return minMoves; 
    }
};
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