<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:
    int shortestPathAllKeys(std::vector<std::string> &grid) {
        
        int nodeCounter = 0;
        std::vector<std::pair<int, int>> nodeCoor(13);
        for (int i = 0; i < grid.size(); ++i){
            for (int j = 0; j < grid[0].size(); ++j)
                if (grid[i][j] != '.' && grid[i][j] != '#'){
                    nodeCoor[getId(grid[i][j])] = {i, j};
                    ++nodeCounter;
                }
        };
        nodeCoor.resize(nodeCounter);

        std::vector<std::vector<std::pair<int, int>>> adjList = buildAdjList(nodeCoor, grid);

        return dijkstraFromZero(adjList);
    }
private:
    const std::vector<std::pair<int, int>> DIRS{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    inline int getId(const char &cell) const {
        if (cell == '@')
            return 0;
        else if (std::islower(cell))
            return (cell - 'a') * 2 + 1;
        else if (std::isupper(cell))
            return (cell - 'A') * 2 + 2;
        return -1;
    }
    std::vector<std::vector<std::pair<int, int>>> buildAdjList(
        const std::vector<std::pair<int, int>> &nodeCoor, const std::vector<std::string> &grid){
        
        std::vector<std::vector<std::pair<int, int>>> result(nodeCoor.size());
        for (int i = 0; i < nodeCoor.size(); ++i)
            result[i] = bfs(nodeCoor[i], grid);
        
        return result;
    }
    std::vector<std::pair<int, int>> bfs(const std::pair<int, int> &coor, const std::vector<std::string> &grid){

        std::vector<std::pair<int, int>> result;
        result.reserve(12);

        std::queue<std::pair<int, int>> que;
        std::vector<std::vector<bool>> visited(grid.size(), std::vector<bool>(grid[0].size()));
        int dist = 0;
        visited[coor.first][coor.second] = true;
        que.push(coor);
        
        while (!que.empty()){
            int qCnt = que.size();
            ++dist;
            while (qCnt--){
                for (const auto &dir : DIRS){
                    int nx = que.front().first + dir.first;
                    int ny = que.front().second + dir.second;
                    if (nx >= 0 && nx < grid.size() && ny >= 0 && ny < grid[0].size()){
                        if (!visited[nx][ny] && grid[nx][ny] != '#'){
                            visited[nx][ny] = true;

                            if (grid[nx][ny] != '.')
                                result.emplace_back(getId(grid[nx][ny]), dist);

                            if (!std::isupper(grid[nx][ny]))
                                que.emplace(nx, ny);
                        }
                    }
                }
                que.pop();
            }
        }

        return result;
    }
    int dijkstraFromZero(const std::vector<std::vector<std::pair<int, int>>> &adjList){
        // struct NodeInfo {
        //     int dist, id;
        //     std::bitset<6> keys;
        //     NodeInfo (const int &d, const int &i): dist(d), id(i) { }
        //     NodeInfo (int &&d, int &&i): dist(std::move(d)), id(std::move(i)) { }
        //     bool operator<(const NodeInfo &other) const {
        //         return this->dist < other.dist;
        //     }
        // };

        using PairInt = std::pair<int, int>;
        const int KEY_MASK = (1 << (adjList.size() - 1) / 2) - 1;
        
        std::priority_queue<PairInt, std::vector<PairInt>, std::greater<PairInt>> pq;
        std::unordered_set<int> visited;
        auto toKey = [](const int &id, const int &keys)->int const {
            return id << 6 | keys;
        };

        pq.emplace(0, 0);
        
        while (!pq.empty()){
            auto now = pq.top();
            pq.pop();
            
            if (visited.contains(now.second))
                continue;
            visited.emplace(now.second);

            auto &dist = now.first;
            int id = now.second >> 6;
            int keys = now.second & KEY_MASK;
            if (keys == KEY_MASK)
                return dist;

            for (const auto &[nb, nextDist] : adjList[id]){
                if (nb & 1){
                    int keyPos = nb >> 1;
                    if ((keys & (1 << keyPos)) == 0)
                        pq.emplace(dist + nextDist, toKey(nb, keys | (1 << keyPos)));
                }
                else if (nb != 0){
                    int keyPos = (nb - 1) >> 1;
                    if ((keys & (1 << keyPos)))
                        pq.emplace(dist + nextDist, toKey(nb, keys));
                }
                else
                    pq.emplace(dist + nextDist, toKey(nb, keys));
            }
            
        }

        return -1;
    }
};
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