<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
    {
        size_t I;
        size_t J;
        int    Moves;
        int    KeysMask;
        
        inline bool HasKey(char c) { return KeysMask & GetBitForKey(c);}
        inline void SetKey(char c) { KeysMask |= GetBitForKey(c);}

        int KeyCount()
        {
            int mask = KeysMask;
            int count = 0;
            while(mask)
            {
                if (mask % 2 == 1)
                    ++count;
                mask /= 2;
            }
            return count;
        }

    private:
        inline int GetBitForKey(char c) { return 1 << (c - 'a');}
    };
    
    int n;
    int m;
    unordered_map<int, vector<bool>> KeysMaskToStatus;

    vector<bool>::reference GetStatus(int KeysMask, size_t i, size_t j)
    { 
        auto & Status = KeysMaskToStatus[KeysMask];
        if (Status.empty())
            Status.resize(n * m);
        return Status[i *m + j]; 
    }

    inline bool IsKey(char c) {return islower(c);}
    inline bool IsLock(char c) {return isupper(c);}

    inline bool IsLocked(State & state, char c){ return IsLock(c) && !state.HasKey(c-'A'+'a');}

    auto GetAdj(size_t i, size_t j){ return array<pair<size_t, size_t>, 4> {make_pair(i, j+1), make_pair(i, j-1), make_pair(i+1, j),  make_pair(i-1, j)}; }
    
    int shortestPathAllKeys(vector<string>& grid) {
        n = grid.size();
        m = grid[0].size();

        queue<State> Q;

        int keys = 0;

        for (size_t i = 0; i < n; ++i)
            for (size_t j = 0; j < m; ++j)
            {
                char c = grid[i][j];
                if (c == '@')
                    Q.push({i, j, 0, 0});
                else
                if (islower(c))
                    ++keys;
            }

        while (!Q.empty())
        {
            auto state = Q.front();
            Q.pop();

            for (auto & [ii, jj] : GetAdj(state.I, state.J))
            {
                if (ii >= n || jj >= m)
                    continue;
                vector<bool>::reference status = GetStatus(state.KeysMask, ii,jj);
                if (status)
                    continue;
                status = true;

                char obj = grid[ii][jj];
                if (obj == '#' || IsLocked(state, obj))
                    continue;

                State nextState {ii, jj, state.Moves + 1, state.KeysMask};
                if (IsKey(obj))
                {
                    nextState.SetKey(obj);
                    if (nextState.KeyCount() == keys)
                        return nextState.Moves;
                }
                Q.push(nextState);
            }
        }
        return -1;
    }
};
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