<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>> dir{ {-1, 0}, {0, 1}, {1, 0}, {0, -1} };

    int shortestPathAllKeys(vector<string>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<vector<bool>>> vis(n, vector<vector<bool>>(m, vector<bool>(64, false))); // 6 possible keys, hence mask size 64
        int keys = 0;
        int start_row, start_col;

        // Find start position and count the number of keys
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == '@') {
                    start_row = i;
                    start_col = j;
                } else if (grid[i][j] >= 'a' && grid[i][j] <= 'z') {
                    keys++;
                }
            }
        }

        queue<vector<int>> q;
        q.push({start_row, start_col, 0, 0}); // row, col, mask, steps
        vis[start_row][start_col][0] = true;

        while (!q.empty()) {
            int row = q.front()[0];
            int col = q.front()[1];
            int mask = q.front()[2];
            int steps = q.front()[3];
            q.pop();

            // Check if we've collected all keys
            if (mask == (1 << keys) - 1) return steps;

            // Explore in all 4 directions
            for (int i = 0; i < 4; i++) {
                int new_row = row + dir[i][0];
                int new_col = col + dir[i][1];

                if (isValid(new_row, new_col, n, m) && grid[new_row][new_col] != '#') {
                    char cell = grid[new_row][new_col];

                    if (cell == '.' || cell == '@') {
                        // Empty or starting cell, continue with the same mask
                        if (!vis[new_row][new_col][mask]) {
                            vis[new_row][new_col][mask] = true;
                            q.push({new_row, new_col, mask, steps + 1});
                        }
                    } else if (cell >= 'a' && cell <= 'z') {
                        // Found a key, update the mask
                        int new_mask = mask | (1 << (cell - 'a'));
                        if (!vis[new_row][new_col][new_mask]) {
                            vis[new_row][new_col][new_mask] = true;
                            q.push({new_row, new_col, new_mask, steps + 1});
                        }
                    } else if (cell >= 'A' && cell <= 'Z') {
                        // Found a lock, check if we have the corresponding key
                        if (mask & (1 << (cell - 'A'))) {
                            if (!vis[new_row][new_col][mask]) {
                                vis[new_row][new_col][mask] = true;
                                q.push({new_row, new_col, mask, steps + 1});
                            }
                        }
                    }
                }
            }
        }

        return -1; // If no solution, return -1
    }

    bool isValid(int row, int col, int n, int m) {
        return row >= 0 && row < n && col >= 0 && col < m;
    }
};

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