<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) {
        int n = grid.size();
        int m = grid[0].size();
        int countKeys = 0;
        queue<tuple<int, int, int, int>> q; // (steps, x, y, keys)
        vector<vector<vector<bool>>> visited(n, vector<vector<bool>>(m, vector<bool>(64, false))); // 64 is used to store states for all key combinations
        
        // 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] >= 'a' && grid[i][j] <= 'z') {
                    ++countKeys;
                }
                if (grid[i][j] == '@') {
                    q.push({0, i, j, 0});
                    visited[i][j][0] = true;
                }
            }
        }
        
        vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        
        while (!q.empty()) {
            auto [steps, x, y, keys] = q.front();
            q.pop();
            
            // Check if we have collected all keys
            if (countKeys == __builtin_popcount(keys)) return steps;
            
            for (auto dir : directions) {
                int newX = x + dir[0];
                int newY = y + dir[1];
                
                if (newX >= 0 && newX < n && newY >= 0 && newY < m && !visited[newX][newY][keys]) {
                    char cell = grid[newX][newY];
                    
                    if (cell == '#') continue; // Wall
                    if (cell >= 'a' && cell <= 'z') { // Key
                        int newKeys = keys | (1 << (cell - 'a'));
                        q.push({steps + 1, newX, newY, newKeys});
                        visited[newX][newY][newKeys] = true;
                    } else if (cell >= 'A' && cell <= 'Z') { // Door
                        if (keys & (1 << (cell - 'A'))) { // Have the key
                            q.push({steps + 1, newX, newY, keys});
                            visited[newX][newY][keys] = true;
                        }
                    } else { // Empty space
                        q.push({steps + 1, newX, newY, keys});
                        visited[newX][newY][keys] = true;
                    }
                }
            }
        }
        return -1;
    }
};
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