<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>

struct T {
  int i;
  int j;
  int keys;  // the keys in the bitmask
};

class Solution {
 public:
  int shortestPathAllKeys(vector<string>& grid) {
    constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int m = grid.size();
    const int n = grid[0].length();
    const int keysCount = getKeysCount(grid);
    const int kKeys = (1 << keysCount) - 1;
    const vector<int> start = getStart(grid);
    int ans = 0;
    queue<T> q{{{start[0], start[1], 0}}};
    vector<vector<vector<bool>>> seen(
        m, vector<vector<bool>>(n, vector<bool>(kKeys)));
    seen[start[0]][start[1]][0] = true;

    while (!q.empty()) {
      ++ans;
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [i, j, keys] = q.front();
        q.pop();
        for (const auto& [dx, dy] : dirs) {
          const int x = i + dx;
          const int y = j + dy;
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          const char c = grid[x][y];
          if (c == '#')
            continue;
          const int newKeys = 'a' <= c && c <= 'f' ? keys | 1 << c - 'a' : keys;
          if (newKeys == kKeys)
            return ans;
          if (seen[x][y][newKeys])
            continue;
          if ('A' <= c && c <= 'F' && (newKeys >> c - 'A' & 1) == 0)
            continue;
          q.emplace(x, y, newKeys);
          seen[x][y][newKeys] = true;
        }
      }
    }

    return -1;
  }

 private:
  int getKeysCount(const vector<string>& grid) {
    int count = 0;
    for (const string& s : grid)
      count += ranges::count_if(s, [](char c) { return 'a' <= c && c <= 'f'; });
    return count;
  }

  vector<int> getStart(const vector<string>& grid) {
    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid[0].length(); ++j)
        if (grid[i][j] == '@')
          return {i, j};
    throw;
  }
};
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