<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}, {0, -1}, {-1, 0}, {1, 0}};
    
    int shortestPathAllKeys(vector<string>& grid) {
        queue<vector<int>> q; // i, j, step, key_status
        int key = 0;
        int n = grid.size();
        int m = grid[0].size();
        
        // Initialize the queue and count the number of keys
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '@') {
                    q.push({i, j, 0, 0}); // Start position with 0 steps and no keys
                } else if (grid[i][j] >= 'a' && grid[i][j] <= 'f') {
                    key++;
                }
            }
        }
        
        int totalBit = (1 << key) - 1;
        // Initialize the visited array
        vector<vector<vector<int>>> vis(n, vector<vector<int>>(m, vector<int>(totalBit + 1, 0)));
        
        while (!q.empty()) {
            vector<int> t = q.front();
            q.pop();
            int i = t[0];
            int j = t[1];
            int step = t[2];
            int bitSt = t[3];
            
            // If we've collected all keys, return the number of steps
            if (bitSt == totalBit) return step;
            
            for (auto dir : dirs) {
                int newi = i + dir[0];
                int newj = j + dir[1];
                
                if (newi >= 0 && newj >= 0 && newi < n && newj < m && grid[newi][newj] != '#') {
                    char ch = grid[newi][newj];
                    int newBitSt = bitSt;
                    
                    if (ch >= 'a' && ch <= 'f') {
                        // Current cell contains a key
                        newBitSt |= (1 << (ch - 'a'));
                    }
                    
                    if (ch >= 'A' && ch <= 'F') {
                        // Current cell contains a lock
                        if ((bitSt & (1 << (ch - 'A'))) == 0) {
                            continue; // Skip if the lock cannot be opened
                        }
                    }
                    
                    if (!vis[newi][newj][newBitSt]) {
                        vis[newi][newj][newBitSt] = 1;
                        q.push({newi, newj, step + 1, newBitSt});
                    }
                }
            }
        }
        
        return -1; // If no path exists to collect all keys
    }
};

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