<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 total;
    vector<int>row={0,0,1,-1};
    vector<int>col={1,-1,0,0};
    vector<vector<vector<int>>> vis;
    int shortestPathAllKeys(vector<string>& grid) {
        // int keys=0;
        int n=grid.size();
        int m=grid[0].size();
        vis.clear();
        vis.resize(n, vector<vector<int>>(m, vector<int>(64,0)));
        total=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]>='A'&&grid[i][j]<='Z')total++;
            }
        }
        total = (int)(pow(2, total)-1);

        queue<pair<pair<int,int>, pair<int,int>>> q;
        for(int x=0;x<m*n;x++){
            // for(int j=0;j<m;j++){
            int i=x/m;
            int j=x%m;
                if(grid[i][j]=='@'){
                    vis[i][j][0]=1;
                    q.push({{i,j}, {0, 0}});
                    break;
                }
            // }
        }

        while(!q.empty()){
            pair<int,int> pos= q.front().first;
            pair<int,int> data= q.front().second;
            q.pop(); int i=pos.first;int j=pos.second; int key = data.first; int len=data.second;
            // cout<<i<<" "<<j<<" "<<" "<<key<<" "<<len<<endl;
            for(int k=0;k<4;k++){
                i+=row[k];j+=col[k];
                if(i>=0 && i<n && j>=0 && j<m){
                    if(grid[i][j]!='#' && vis[i][j][key]==0){
                        if(grid[i][j]>='a' && grid[i][j]<='z'){//key
                            int lock=grid[i][j]-'a';
                            // cout<<"key "<<lock<<endl;
                            // key = key | 1<<lock;
                            if((key | 1<<lock)==total) return len+1;
                            q.push({{i,j}, {(key | 1<<lock), len+1}});
                            // cout<<"->"<<i<<" "<<j<<" "<<" "<<key<<" "<<len<<endl;
                            vis[i][j][key]=1;
                        }
                        else if(grid[i][j]>='A' && grid[i][j]<='Z'){//lock
                            int lock=grid[i][j]-'A';
                            if(((key>>lock) & 1)>0){
                                // cout<<"lock "<<lock<<endl;
                                q.push({{i,j}, {key, len+1}});
                            // cout<<"->"<<i<<" "<<j<<" "<<" "<<key<<" "<<len<<endl;

                                vis[i][j][key]=1;
                            }
                        }
                        else{
                            q.push({{i,j}, {key, len+1}});
                            // cout<<"->"<<i<<" "<<j<<" "<<" "<<key<<" "<<len<<endl;

                            vis[i][j][key]=1;
                        }
                    }
                }
                i-=row[k];j-=col[k];
            }
        }

        return -1;
    }


};
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