<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:
    void find(int i,int j,vector<string>&grid,string &s,
              vector<vector<unordered_set<string>>>&vis,
              queue< pair <  pair<int,int>, string  > >&qmt,int m,int n)
    {
        if(i>=m||i<0||j>=n||j<0)return;
        if(grid[i][j]=='#')return;
        

        
        if(grid[i][j]>='a'&&grid[i][j]<='z')
        {
            string tmp=s;
            
            int flg=0;
            for(auto&p:tmp)if(p==grid[i][j])flg=1;
            if(flg==0)
            tmp.push_back(grid[i][j]);
            sort(tmp.begin(),tmp.end());
            if(vis[i][j].find(tmp)!=vis[i][j].end())return;
            vis[i][j].insert(tmp);
            qmt.push({ {i,j},tmp});
            return;
        }
        if(grid[i][j]=='.'||grid[i][j]=='@')
        {
            string tmp=s;
            if(vis[i][j].find(tmp)!=vis[i][j].end())return;
            vis[i][j].insert(tmp);
            qmt.push({ {i,j},tmp});
            return;
        }
        if(grid[i][j]>='A'&&grid[i][j]<='Z')
        {
            string tmp=s;
            char t=grid[i][j]+32;
            int flg=0;
            for(auto&p:tmp)if(p==t)flg=1;
            if(flg==0)return;
            
            if(vis[i][j].find(tmp)!=vis[i][j].end())return;
            vis[i][j].insert(tmp);
            qmt.push({ {i,j},tmp});
        }
    }
    
    int shortestPathAllKeys(vector<string>& grid) 
    {
        int m=grid.size(),n=grid[0].size(),cntk=0,strti,strtj;
        vector<vector<unordered_set<string>>>mp
        (m,vector<unordered_set<string>> (n, unordered_set<string> {}));
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]=='@')
                {
                    strti=i;strtj=j;
                }
                if(grid[i][j]>='a'&&grid[i][j]<='z')cntk++;
            }
        }
        int ans=-1;
        queue< pair <  pair<int,int>, string > >qmt;
        qmt.push({{strti,strtj},""});
        while(qmt.size()>0)
        {
            ans++;
            int sz=qmt.size();
            
            for(int k=0;k<sz;k++)
            {
            pair< pair<int,int>,string >p=qmt.front();
            qmt.pop();
            int i=p.first.first,j=p.first.second;
            string s=p.second;

            if(s.size()==cntk)return ans;
            find(i+1,j,grid,s,mp,qmt,m,n);
            find(i-1,j,grid,s,mp,qmt,m,n);
            find(i,j+1,grid,s,mp,qmt,m,n);
            find(i,j-1,grid,s,mp,qmt,m,n);
            }
        }
        return -1;
    }
};
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