<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(vector<string>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int sr ;
        int sc ;
        int cnt = 0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j] == '@'){
                    sr=i;
                    sc=j;
                }
                if(grid[i][j]>='a' && grid[i][j]<='f') cnt++;
            }
        }
        priority_queue<pair<pair<int,set<vector<int>>>,pair<pair<int,int>,int>>,
        vector<pair<pair<int,set<vector<int>>>,pair<pair<int,int>,int>>>,
        greater<pair<pair<int,set<vector<int>>>,pair<pair<int,int>,int>>>>q;
        set<vector<int>>vss;
        q.push({{0,vss},{{sr,sc},0}});
        int dr[4] = {0,0,1,-1};
        int dc[4] = {1,-1,0,0};
        while(!q.empty()){
            auto dis = q.top().first.first;
            auto vis = q.top().first.second;
            auto r = q.top().second.first.first;
            auto c = q.top().second.first.second;
            auto keys = q.top().second.second;
            // cout<<r<<" "<<c<<" "<<keys<<" "<< endl;
            q.pop();
            if(vss.find({r,c,keys})!=vss.end()) continue;
            vss.insert({r,c,keys});
            if (keys == (1 << cnt) - 1) return dis;
            for(int i=0;i<4;i++){
                int nr = r+dr[i];
                int nc = c+dc[i];

                if(nr>=0 && nr<n && nc>=0 && nc<m){
                    if(grid[nr][nc]=='#') continue;
                    if(grid[nr][nc]>='a' && grid[nr][nc]<='f'){
                        int num =  grid[nr][nc]-'a';
                        int nkeys = keys|(1<<num);
                        if(nkeys==( (1<<cnt)-1 ) ) return dis+1 ;
                        q.push({{dis+1,vis},{{nr,nc},nkeys}});
                        // q.push({dis+1,{{nr,nc},keys}});
                    }
                    else if(grid[nr][nc]>='A' && grid[nr][nc]<='F'){
                        int num = grid[nr][nc] - 'A';
                        if((keys&(1<<num)) !=0){
                        q.push({{dis+1,vis},{{nr,nc},keys}});
                        }
                    }
                    else if(grid[nr][nc]=='.' || grid[nr][nc]=='@'){
                        q.push({{dis+1,vis},{{nr,nc},keys}});
                    }
                }
            }
            // return -1;
        }
        return -1;
    }
};
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