<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:
    #define pii pair<int,pair<vector<int>,pair<int,int>>>
    class comp{
        public:
        bool operator()(pii a,pii b){
            auto v1 = a.second.first;
            auto v2 = b.second.first;

            int c1= 0,c2=0;
            for(auto i:v1)c1+=i;
            for(auto i:v2)c2+=i;

            if(c1 == c2){
                return a.first > b.first;
            }

            return c1 < c2;
        }
    };

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

        keys.assign(6,0);

        int locks = 0;

        queue<vector<int>> q;

        int dx[]=  {-1,0,1,0};
        int dy[] = {0,-1,0,1};

        for(int i = 0;i < n;i++){
            for(int j = 0;j < m;j++){
                if(grid[i][j] == '@'){
                    q.push({i,j,0,0});
                }
                if((grid[i][j] >= 'a' && grid[i][j] <= 'f')){
                    locks++;
                }
            }
        }
        int tg = (1 << locks) - 1;

        map<vector<int>,int> vis;

        while(!q.empty()){
            auto fnode = q.front();
            q.pop();

            int x = fnode[0];
            int y = fnode[1];
            int moves = fnode[2];
            auto key = fnode[3];

            if(key == tg){
                return moves;
            }

            if(vis[{x,y,key}])continue;

            vis[{x,y,key}] = 1;

            for(int d = 0;d < 4;d++){
                int nx = x + dx[d];
                int ny = y + dy[d];

                if(nx >= 0 && ny >= 0 && nx < n && ny < m && grid[nx][ny] != '#' && !vis[{nx,ny,key}]){
                    if(grid[nx][ny] == '.' || grid[nx][ny] == '@'){
                        q.push({nx,ny,1 + moves,key});
                    }
                    else
                    {
                        char ch = grid[nx][ny];
                        if(ch >= 'A' && ch <= 'F'){
                            int nn = grid[nx][ny] - 'A';
                            if((key >> nn) & 1){
                                q.push({nx,ny,1 + moves,key});
                            }
                        }
                        else if(ch >= 'a' && ch <= 'f'){
                            int nn = grid[nx][ny] - 'a';
                            int nk = key | (1 << nn);
                            // if(!vis[{nx,ny,nk}]){
                                // vis[{nx,ny,nk}] = 1;
                                q.push({nx,ny,1 + moves,nk});
                            // }
                        }
                    }
                }
            }
        }
        return -1;
    }
};
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