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

int shortestPathAllKeys(vector<string>& grid) {
    int n = grid.size();
    int m = grid[0].size();
    int key = 0;
    int startX, startY;

    // Find the total number of keys and the starting position
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (grid[i][j] >= 'a' && grid[i][j] <= 'z') {
                key++;
            }
            if (grid[i][j] == '@') {
                startX = i;
                startY = j;
            }
        }
    }

    // BFS Initialization
    queue<pair<pair<int, int>, int>> q; // ((x, y), collectedKeys bitmask)
    q.push({{startX, startY}, 0});
    unordered_set<string> visited;  // Track visited states
    visited.insert(to_string(startX) + "," + to_string(startY) + ",0");
    
    int step = 0;
    
    while (!q.empty()) {
        int sz = q.size();
        while (sz--) {
            int x = q.front().first.first;
            int y = q.front().first.second;
            int keys = q.front().second;
            q.pop();

            // If all keys have been collected
            if (keys == (1 << key) - 1) {
                return step;
            }

            for (auto d : dir) {
                int nx = x + d[0];
                int ny = y + d[1];
                int newKeys = keys;

                if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
                    char cell = grid[nx][ny];

                    if (cell == '#') continue; // Wall

                    if (cell >= 'a' && cell <= 'z') { // It's a key
                        newKeys |= (1 << (cell - 'a')); // Collect the key
                    }

                    if (cell >= 'A' && cell <= 'Z') { // It's a door
                        if ((newKeys & (1 << (cell - 'A'))) == 0) {
                            continue; // Skip if we don't have the key
                        }
                    }

                    string state = to_string(nx) + "," + to_string(ny) + "," + to_string(newKeys);
                    if (visited.find(state) == visited.end()) {
                        visited.insert(state);
                        q.push({{nx, ny}, newKeys});
                    }
                }
            }
        }
        step++;
    }
    
    return -1; // No path found
}

};
59010