<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>

#define ll long long
#define pb push_back
class Solution {
public:
    int shortestPathAllKeys(vector<string>& grid) {
        ll n = grid.size();
        ll m = grid[0].size();
        ll k = 0;
        ll sx =-1;
        ll sy =-1;
        for(ll i=0;i<n;i++){
            for(ll j=0;j<m;j++){
                if(grid[i][j]>='a' && grid[i][j]<='f'){
                    k++;
                   
                }
                if(grid[i][j]=='@'){
                    sx=i;
                    sy=j;
                }
            }
        }
        ll mask = (1<<k)-1;
        vector<vector<vector<ll>>>dist(n,vector<vector<ll>>(m,vector<ll>(mask+1,1e11)));
        queue<vector<ll>>q;
        q.push({0,sx,sy,0});
        dist[sx][sy][0]=0;
        ll delx[4] = {-1,0,1,0};
        ll dely[4] = {0,1,0,-1};
        while(!q.empty()){
            ll xnode = q.front()[1];
            ll ynode = q.front()[2];
            ll distance = q.front()[0];
            ll bitmask = q.front()[3];
            q.pop();
            for(ll i=0;i<4;i++){
                ll newx = xnode + delx[i];
                ll newy = ynode + dely[i];
                if(newx>=0 && newx<n && newy>=0 && newy<m && grid[newx][newy]!='#'){
                    if(grid[newx][newy]>='A' && grid[newx][newy]<='F'){
                        if(((1<<(grid[newx][newy]-'A')) & bitmask)!=0){
                            if(dist[newx][newy][bitmask]>distance+1){
                               dist[newx][newy][bitmask]=distance+1; 
                               q.push({distance+1,newx,newy,bitmask});
                            }
                        }
                    }else if(grid[newx][newy]>='a' && grid[newx][newy]<='f'){
                        ll new_mask = (bitmask | (1<<(grid[newx][newy]-'a')));
                        if(dist[newx][newy][new_mask]>distance+1){
                            dist[newx][newy][new_mask]=distance+1;
                            q.push({distance+1,newx,newy,new_mask}); 
                        }
                    }
                    else{
                            if(dist[newx][newy][bitmask]>distance+1){
                               dist[newx][newy][bitmask]=distance+1; 
                               q.push({distance+1,newx,newy,bitmask});
                            }
                    }
                }
            }

        }

        ll mini = 1e9;
        for(ll i=0;i<n;i++){
            for(ll j=0;j<m;j++){
                mini=min(mini,dist[i][j][mask]);
            }
        }

        if(mini==1e9)mini=-1;
        return mini;   
    }
};
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