<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:
    map<array<int, 3>, int> odw;
    queue<array<int, 3>> order;
    int n, m;
    vector<string> grid;
    bool gd(int x, int y){
        if(x<0||y<0||x>=n||y>=m||grid[x][y]=='#') return false;
        return true;
    }
    void gen(int x, int y, int mask, int last1, int last2){
        if(!gd(x, y)) return;
        int new_mask=mask;
        if(grid[x][y]>='a'&&grid[x][y]<='z'){
            int pm=grid[x][y]-'a';
            new_mask|=(1<<pm);
        }
        if(grid[x][y]>='A'&&grid[x][y]<='Z'){
            int pm=grid[x][y]-'A';
            if(!(new_mask&(1<<pm))) return;
        }
        if(odw.find({x, y, new_mask})!=odw.end()) return;
        odw[{x, y, new_mask}]=odw[{last1, last2, mask}]+1;
        order.push({x, y, new_mask});
    }
    int shortestPathAllKeys(vector<string>& z) {
        grid=z;
        odw.clear();
        while(!order.empty()) order.pop();
        n=grid.size();
        m=grid[0].size();
        int k=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]=='@'){
                    order.push({i, j, 0});
                    odw[{i, j, 0}]=0;
                }
                if(grid[i][j]>='a'&&grid[i][j]<='z') k++;
            }
        }
        int target=(1<<k)-1;
        while(!order.empty()){
            auto x = order.front();
            order.pop();
            gen(x[0]-1, x[1], x[2], x[0], x[1]);
            gen(x[0]+1, x[1], x[2], x[0], x[1]);
            gen(x[0], x[1]-1, x[2], x[0], x[1]);
            gen(x[0], x[1]+1, x[2], x[0], x[1]);
        }
        int ans=1e9;
        for(auto v : odw){
            if(v.first[2]==target){
                ans=min(ans, v.second);
            }
        }
        return ans==1e9 ? -1 : ans;
    }
};
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