<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 keyCnt = 0;
        std::pair<int, int> source;
        std::vector<std::pair<int, int>> keyCoors(6);
        for (int i = 0; i < grid.size(); ++i){
            for (int j = 0; j < grid[0].size(); ++j)
                if (std::islower(grid[i][j])){
                    keyCoors[grid[i][j] - 'a'] = {i, j};
                    ++keyCnt;
                }
                else if (grid[i][j] == '@')
                    source = {i, j};
        };
        keyCoors.resize(keyCnt);

        return aStarAlgorithm(source, keyCoors, grid);
    }
private:
    int aStarAlgorithm(const std::pair<int, int> &source, const std::vector<std::pair<int, int>> &keyCoors,
        const std::vector<std::string> &grid){
        
        const std::vector<std::pair<int, int>> DIRS{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
        const int KEY_MASK = (1 << keyCoors.size()) - 1;

        auto isInBoundary = [&grid](const int &x, const int &y)->bool const {
            if (x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size())
                return true;
            return false;
        };

        auto hScore = [&keyCoors](const pair<int, int> &coor, const int &keys)->int const {
            int result = INT_MAX;
            for (int i = 0; i < keyCoors.size(); ++i)
                if ((keys & 1 << i) == 0)
                    result = std::min(result, std::abs(coor.first - keyCoors[i].first) + std::abs(coor.second - keyCoors[i].second));
            return result == INT_MAX ? 0 : result;
        };
        
        using NodeInfo = tuple<int, int, pair<int, int>, int>;       // {(fScore, gScore, coordination, keys)}
        struct customComp {
            bool operator()(const NodeInfo &a, const NodeInfo &b) const {
                return std::get<0>(a) > std::get<0>(b);
            }
        };

        std::priority_queue<NodeInfo, std::vector<NodeInfo>, customComp> openList;
        std::unordered_set<int> closedList;
        auto hashing = [](const pair<int, int> &coor, const int &keys)->int const {
            return coor.first << 12 | coor.second << 6 | keys;
        };
        openList.emplace(0 + hScore(source, 0), 0, source, 0);
        
        while (!openList.empty()){
            int gScore = std::get<1>(openList.top());
            auto coor = std::get<2>(openList.top());
            int keys = std::get<3>(openList.top());
            if (keys == KEY_MASK)
                return gScore;
            openList.pop();
            closedList.emplace(hashing(coor, keys));

            for (const auto &dir : DIRS){
                int nx = coor.first + dir.first;
                int ny = coor.second + dir.second;
                if (isInBoundary(nx, ny) && grid[nx][ny] != '#'){
                    if (std::isupper(grid[nx][ny]) && (keys & 1 << (grid[nx][ny] - 'A')) == 0)
                        continue;
                    int newKeys = std::islower(grid[nx][ny]) ? keys | 1 << (grid[nx][ny] - 'a') : keys;
                    int newHashValue = hashing({nx, ny}, newKeys);
                    if (closedList.find(newHashValue) == closedList.end()) {
                        int newGScore = gScore + 1;
                        int newFScore = newGScore + hScore({nx, ny}, newKeys);
                        openList.emplace(newFScore, newGScore, std::make_pair(nx, ny), newKeys);
                    }
                }
            }
        }

        return -1;
    }
};
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