<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<int> dx = {1, -1, 0, 0};
    vector<int> dy = {0, 0, 1, -1};
    
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ct = 0;
        int start_x = 0, start_y = 0;
        queue<vector<int>> q;
        unordered_set<string> visited;
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == '@'){
                    start_x = i;
                    start_y = j;
                }
                if(grid[i][j] >= 'a' && grid[i][j] <= 'f'){
                    ct++;
                }
            }
        }

        int final_mask = (1 << ct) - 1;
        q.push({0, start_x, start_y, 0});
        visited.insert("0 " + to_string(start_x) + " " + to_string(start_y));
        
        while(!q.empty()){
            auto current = q.front();
            q.pop();
            int steps = current[0];
            int x = current[1];
            int y = current[2];
            int mask = current[3];

            if(x < 0 || y < 0 || x >= m || y >= n){
                continue;
            }
            
            char cell = grid[x][y];
            if(cell == '#'){
                continue;
            }
            if(cell >= 'A' && cell <= 'F'){
                int lock = cell - 'A';
                if((mask & (1 << lock)) == 0){
                    continue;
                }
            }
            if(cell >= 'a' && cell <= 'f'){
                int key = cell - 'a';
                mask |= (1 << key);
            }
            if(mask == final_mask){
                return steps;
            }
            
            for(int i = 0; i < 4; i++){
                int nx = x + dx[i];
                int ny = y + dy[i];
                string state = to_string(mask) + " " + to_string(nx) + " " + to_string(ny);
                if(visited.find(state) == visited.end()){
                    q.push({steps + 1, nx, ny, mask});
                    visited.insert(state);
                }
            }
        }
        return -1;
    }
};

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