<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) {
        unordered_map<char, int> keys = {{'a', 0}, {'b', 1}, {'c', 2}, {'d', 3}, {'e', 4}, {'f', 5}};
        unordered_map<char, int> locks = {{'A', 0}, {'B', 1}, {'C', 2}, {'D', 3}, {'E', 4}, {'F', 5}};
        vector<pair<int, int>> dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        set<vector<int>> visit;

        queue<vector<int>> q;
        int key_cnt = 0;
        int m = grid.size();
        int n = grid[0].size();
        for(int i = 0;i< m;i++){
            for(int j = 0;j<n;j++){
                char c = grid[i][j];
                if(c == '@'){
                    q.push({i,j, 0, 0});
                } else if(keys.count(c))
                    key_cnt++;
            }
        }

        while(!q.empty()){
            vector<int> tpl = q.front(); q.pop();
            int i = tpl[0], j = tpl[1], found=tpl[2], step = tpl[3];
            char c = grid[i][j];

            if(locks.count(c) && !((found>>locks[c]) & 1)) // c is lock, but the lock key haven't been found
                continue;

            if(c == '#')
                continue;

            if(keys.count(c)){
                found |= (1<<keys[c]);
                if(found == ((1<<key_cnt)-1))
                    return step;
            }

            for(pair<int,int> d: dir) {
                int x = i+d.first, y = j+d.second;
                if(x>=0&&x<m&&y>=0&&y<n&&grid[x][y]!='#'){
                    vector<int> p = {x, y, found};
                    if(!visit.count(p)){
                        visit.insert(p);
                        q.push({x, y, found, step+1});
                    }
                }
            }
        }
        return -1;
    }
};
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