<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 {
private:
    pair <int, int> getStart(vector<string>& grid, int &maxlen){
        int m = grid.size();
        int n = grid[0].size();
        pair<int, int> ans;

        for(int i=0; i<m; i++){ 
            for(int j=0; j<n; j++){
                if(grid[i][j]=='@') ans = {i,j};
                if (grid[i][j]>='a' && grid[i][j]<='f') maxlen=max(maxlen, grid[i][j]-'a' + 1);
            }
        }
        return ans;
    }

public:
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size();
        int n = grid[0].size();

        queue <pair <int, pair <int, int>>> q;
        int maxlen = 0;
        pair <int, int> p = getStart(grid, maxlen);
        q.push({0, p});
        cout<<maxlen<<endl;
        cout<<p.first<<' '<<p.second<<endl;

        set <vector <int>> vis;
        vis.insert({0, p.first, p.second});
        int dx[] = {0, 0, -1, 1};
        int dy[] = {-1, 1, 0, 0};
        int lvl = 0;

        while(!q.empty()){
            int sz = q.size();
            while(sz--){
                auto keys = q.front().first;
                int r = q.front().second.first;
                int c = q.front().second.second;
                q.pop();

                if (keys == ((1<<maxlen)-1)) return lvl;
                
                for(int k=0; k<4; k++){
                    int new_keys = keys;
                    int x = r + dx[k];
                    int y = c + dy[k];

                    if (x<0 || y<0 || x==m || y==n) continue;
                    if (grid[x][y] == '#') continue;
                    if (grid[x][y] >= 'a' && grid[x][y] <= 'f'){
                        new_keys |= (1<<(grid[x][y] - 'a'));
                    }else if(grid[x][y]>='A' && grid[x][y]<='F'){
                        if ((new_keys & (1<<(grid[x][y] - 'A'))) == 0){
                            continue;
                        }
                    }

                    if (vis.find({new_keys, x, y}) != vis.end()) continue;
                    q.push({new_keys, {x, y}});
                    vis.insert({new_keys, x, y});
                }
            }
            lvl++;
        }

        return -1;
    }
};
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