<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:

struct State {
    int x, y, keys, steps;
    State(int _x, int _y, int _keys, int _steps) : x(_x), y(_y), keys(_keys), steps(_steps) {}
};

string toKey(int x, int y, int keys) {
    return to_string(x) + "," + to_string(y) + "," + to_string(keys);
}

int shortestPathAllKeys(vector<string>& grid) {
    int m = grid.size(), n = grid[0].size();
    int start_x, start_y, num_keys = 0;

    // Initialize start position and count keys
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] == '@') {
                start_x = i;
                start_y = j;
            } else if (islower(grid[i][j])) {
                num_keys++;
            }
        }
    }

    int all_keys = (1 << num_keys) - 1;  // All keys bitmask
    queue<State> q;
    unordered_set<string> visited;

    q.push(State(start_x, start_y, 0, 0));
    visited.insert(toKey(start_x, start_y, 0));

    // BFS
    while (!q.empty()) {
        State current = q.front();
        q.pop();

        int x = current.x, y = current.y, keys = current.keys, steps = current.steps;
        for (auto [dx, dy] : vector<pair<int, int>>{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}) {
            int nx = x + dx, ny = y + dy;
            if (nx < 0 || ny < 0 || nx >= m || ny >= n) continue;

            char cell = grid[nx][ny];
            if (cell == '#') continue;  // Wall

            int new_keys = keys;
            if (islower(cell)) {
                new_keys |= (1 << (cell - 'a'));  // Collect key
            }

            if (isupper(cell) && !(keys & (1 << (cell - 'A')))) {
                continue;  // Lock and we don't have the key
            }

            if (new_keys == all_keys) return steps + 1;  // Collected all keys

            string state_key = toKey(nx, ny, new_keys);
            if (!visited.count(state_key)) {
                visited.insert(state_key);
                q.push(State(nx, ny, new_keys, steps + 1));
            }
        }
    }

    return -1;  // Impossible to collect all keys
}

};
39503