<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>

int dr[] = {-1,0,1,0};
int dc[] = {0,1,0,-1};

class Solution {

    bool keyAvailable(char lock, string &s){
        if(s.find('a'+(lock-'A')) != string::npos) return true;
        return false;
    }

    void pickupKey(char key, string &s){
        if(s.find(key) == string::npos){
            s += key;
            sort(s.begin(),s.end());
        }
    }

public:
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int>>dist(m,vector<int>(n,1e9));
        vector<vector<set<string>>>maxKeys(m,vector<set<string>>(n));
        //
        priority_queue<pair<pair<int,string>,vector<int>>,vector<pair<pair<int,string>,vector<int>>>,greater<pair<pair<int,string>,vector<int>>>>pq;
        int x = 0, y = 0;
        int tot_keys = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == '@'){
                    x = i; y = j;
                }
                if(grid[i][j] >= 'a' && grid[i][j] <= 'z'){
                    tot_keys++;
                }
            }
        }
        pq.push({{0,""},{x,y}});
        dist[x][y] = 0;
        int res = 1e9;
        while(!pq.empty())
        {
            auto p = pq.top(); pq.pop();
            int step = p.first.first, r = p.second[0], c = p.second[1];
            string keys = p.first.second;
            if(keys.size() == tot_keys) res = min(res, step);
            for(int i=0; i<4; i++)
            {
                int nr = r + dr[i], nc = c + dc[i];
                if(nr<0 || nr>=m || nc<0 || nc>=n || grid[nr][nc]=='#') continue;
                string newKeys = keys;
                if(grid[nr][nc] >= 'A' && grid[nr][nc] <= 'Z')
                {
                    if(keyAvailable(grid[nr][nc],keys) == false) continue;
                }
                else if(grid[nr][nc] >= 'a' && grid[nr][nc] <= 'z')
                {
                    pickupKey(grid[nr][nc],newKeys);
                }
                if((dist[nr][nc] == 1e9) ||
                (maxKeys[nr][nc].find(newKeys) == maxKeys[nr][nc].end()))
                {
                    dist[nr][nc] = step + 1;
                    maxKeys[nr][nc].insert(newKeys);
                    pq.push({{step+1,newKeys},{nr,nc}});
                }
            }
        }
        if(res != 1e9) return res;
        return -1;
    }
};
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