<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 isLower(char& c){
        return c>='a' && c<='f';
    }
    
    int isUpper(char& c){
        return c>='A' && c<='F';
    }

    int shortestPathAllKeys(vector<string>& grid) {
        int R = grid.size(); int C = grid[0].size();
        int KS = 0;
        //                                                key
        unordered_map<int, unordered_map<int, unordered_set<int>>> M;
        //       x, y, dist, keysCombo
        queue<vector<int>> Q;
        vector<vector<int>> D = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};

        for(int i=0; i<R && Q.empty(); i++) {
            for(int j=0; j<C && Q.empty(); j++){
                if(grid[i][j] == '@'){
                    Q.push({i, j, 0, 0});
                    M[i][j].insert(0);
                    break;
                }
            }
        }

        for(int i=0; i<R; i++) {
            for(int j=0; j<C; j++){
                if(isLower(grid[i][j])){
                    KS = (KS | (1<<(grid[i][j]-'a')));
                }
            }
        }

        // cout<<KS<<endl;
        while(!Q.empty()) {
            vector<int> E = Q.front(); Q.pop();
            int x = E[0]; int y = E[1];
            int d = E[2]; int k = E[3];

            if(k == KS)
                return d;
            
            for(auto& dir: D){
                int i = dir[0] + x; int j = dir[1] + y;
                if(i>=0 && j>=0 && i<R && j<C && grid[i][j]!='#'){
                    // cout<<i<<" "<<j<<" "<<k<<endl;
                    int newk = (isLower(grid[i][j]))?(k | (1<<(grid[i][j]-'a'))):-1; 
                    int lk = (isUpper(grid[i][j]))?(1<<(grid[i][j]-'A')):-1;
                    if((grid[i][j] == '.') && (M[i][j].find(k) == M[i][j].end())) {
                        Q.push({i, j, d+1, k}); M[i][j].insert(k);
                    } else if(isLower(grid[i][j]) && (M[i][j].find(newk) == M[i][j].end())) {
                        Q.push({i, j, d+1, newk}); M[i][j].insert(newk);
                    } else if((lk & k) && (M[i][j].find(k) == M[i][j].end())) {
                        Q.push({i, j, d+1, k}); M[i][j].insert(k);
                    }
                }
            }
        }

        return -1;
    }
};
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