<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 m=grid.size();
        int n=grid[0].size();
        int ans=0;
        int count=0;
        unordered_map<char,int>mp;
        int si,sj;
        for(string s:grid){
            for(char c:s){
                if(c>='a' && c<='z' && mp.find(c)==mp.end()){
                    mp[c]=count;
                    count++;
                }
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]=='@'){
                    si=i;
                    sj=j;
                }
            }
        }
        for(auto itr:mp){
            // cout<<itr.first<<" "<<itr.second<<endl;
            ans=ans|(1<<itr.second);
        }
        vector<vector<vector<int>>>vis(m,vector<vector<int>>(n,vector<int>(64,-1)));
        queue<vector<int>>q;
        q.push({si,sj,0,1});
        // cout<<si<<" "<<sj<<" "<<ans<<endl;
        while(!q.empty()){
            vector<int>temp=q.front();
            q.pop();
            // cout<<temp[0]<<" "<<temp[1]<<" "<<temp[2]<<" "<<temp[3]<<endl;
            
            if(temp[0]<0 || temp[1]<0 || temp[0]==m || temp[1]==n || grid[temp[0]][temp[1]]=='#')continue;
            if(vis[temp[0]][temp[1]][temp[2]]!=-1)continue;
            vis[temp[0]][temp[1]][temp[2]]=temp[3];
            int vit=temp[2];
            int nvit=temp[2];
            if(grid[temp[0]][temp[1]]>='a' && grid[temp[0]][temp[1]]<='z'){
                nvit=nvit|(1<<mp[grid[temp[0]][temp[1]]]);
                if(nvit==ans)return temp[3]-1;
            }
            else if(grid[temp[0]][temp[1]]>='A' && grid[temp[0]][temp[1]]<='Z'){
                // cout<<"this"<<endl;
                char c='a'+(grid[temp[0]][temp[1]]-'A');
                // cout<<c<<endl;
                if((vit&(1<<mp[c]))==0)continue;
            }
            
            q.push({temp[0]+1,temp[1],nvit,temp[3]+1});
            q.push({temp[0]-1,temp[1],nvit,temp[3]+1});
            q.push({temp[0],temp[1]+1,nvit,temp[3]+1});
            q.push({temp[0],temp[1]-1,nvit,temp[3]+1});
            
        }
        return -1;
    }
};
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