<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 {
private:
    bool present(char ch, string keys)
    {
        for(int i=0;i<keys.size();i++)
        {
            if(keys[i]==ch)
                return true;
        }
        return false;
    }
public:
    int dx[4]={0, 0, 1, -1};
    int dy[4]={1, -1, 0, 0};
    int shortestPathAllKeys(vector<string>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        int keys=0;
        int x, y, c=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]=='@')
                    x=i, y=j;
                if(grid[i][j]>='a' && grid[i][j]<='f')
                    c++;
            }
        }
        queue<pair<pair<int, int>, string>> q;
        q.push({{x, y}, ""});
        unordered_set<string> vis;
        int moves=0;
        while(!q.empty())
        {
            int l=q.size();
            for(int i=0;i<l;i++)
            {
                int row=q.front().first.first;
                int col=q.front().first.second;
                string keys=q.front().second;
                q.pop();
                if(keys.size()==c)
                    return moves;
                string key=to_string(row)+":"+to_string(col)+":"+keys;
                if(vis.find(key)!=vis.end())
                    continue;
                vis.insert(key);
                for(int j=0;j<4;j++)
                {
                    int nrow=row+dx[j];
                    int ncol=col+dy[j];
                    if(nrow<0 || nrow>=n || ncol<0 || ncol>=m || grid[nrow][ncol]=='#')
                        continue;
                    char ch=grid[nrow][ncol];
                    if(ch>='A' && ch<='F' && present(ch+32, keys))
                        q.push({{nrow, ncol}, keys});
                    else if(ch=='.' || ch=='@')
                        q.push({{nrow, ncol}, keys});
                    else if(ch>='a' && ch<='f')
                    {
                        if(present(ch, keys))
                            q.push({{nrow, ncol}, keys});
                        else
                            q.push({{nrow, ncol}, keys+ch});
                    }
                }
            }
            moves++;
        }
        return -1;
    }
};
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