Shortest Path to Get All Keys
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0864.Shortest%20Path%20to%20Get%20All%20Keys/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0864.Shortest%20Path%20to%20Get%20All%20Keys/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0864.Shortest%20Path%20to%20Get%20All%20Keys/images/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>
Bit Manipulation; Breadth-First Search; Array; Matrix
class Solution {
    private int[] dirs = {-1, 0, 1, 0, -1};

    public int shortestPathAllKeys(String[] grid) {
        int m = grid.length, n = grid[0].length();
        int k = 0;
        int si = 0, sj = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                char c = grid[i].charAt(j);
                if (Character.isLowerCase(c)) {
                    // 累加钥匙数量
                    ++k;
                } else if (c == '@') {
                    // 起点
                    si = i;
                    sj = j;
                }
            }
        }
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {si, sj, 0});
        boolean[][][] vis = new boolean[m][n][1 << k];
        vis[si][sj][0] = true;
        int ans = 0;
        while (!q.isEmpty()) {
            for (int t = q.size(); t > 0; --t) {
                var p = q.poll();
                int i = p[0], j = p[1], state = p[2];
                // 找到所有钥匙，返回当前步数
                if (state == (1 << k) - 1) {
                    return ans;
                }
                // 往四个方向搜索
                for (int h = 0; h < 4; ++h) {
                    int x = i + dirs[h], y = j + dirs[h + 1];
                    // 在边界范围内
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        char c = grid[x].charAt(y);
                        // 是墙，或者是锁，但此时没有对应的钥匙，无法通过
                        if (c == '#'
                            || (Character.isUpperCase(c) && ((state >> (c - 'A')) & 1) == 0)) {
                            continue;
                        }
                        int nxt = state;
                        // 是钥匙
                        if (Character.isLowerCase(c)) {
                            // 更新状态
                            nxt |= 1 << (c - 'a');
                        }
                        // 此状态未访问过，入队
                        if (!vis[x][y][nxt]) {
                            vis[x][y][nxt] = true;
                            q.offer(new int[] {x, y, nxt});
                        }
                    }
                }
            }
            // 步数加一
            ++ans;
        }
        return -1;
    }
}