<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 {
    const int inf = 1e9 + 9;
    int dx[4] = {-1 , 1 , 0 , 0};
    int dy[4] = {0 , 0 , -1 , 1};
public:
    int shortestPathAllKeys(vector<string>& grid) {
        int n = grid.size(), m = grid[0].size();
        int start = -1;
        set<char> s;
        for (int i = 0 ; i < n ; i++) {
            for (int j = 0 ; j < m ; j++) {
                if (grid[i][j] == '@') start = i*m+j;
                if (grid[i][j] == '.' || grid[i][j] == '#') continue;
                if (grid[i][j] >= 'a' && grid[i][j] <= 'z') s.insert(grid[i][j]);
            }
        }

        map<char, int> m1;
        int k = 0;
        for (auto it : s) {
            m1[it] = k++;
        }

        queue<pair<int ,int>> q;
        q.push({start, 0});

        vector<vector<int>> visited(n*m, vector<int> ((1 << k)));
        vector<vector<int>> d(n*m, vector<int> ((1 << k), inf));
        
        visited[start][0] = 1;
        d[start][0] = 0;

        while (!q.empty()) {
            pair<int ,int> x = q.front();
            q.pop();

            int i = x.first / m , j = x.first % m , mask = x.second;
            for (int idx = 0 ; idx < 4 ; idx++) {
                int i1 = i + dx[idx], j1 = j + dy[idx];
                if (i1 < 0 || i1 >= n || j1 < 0 || j1 >= m || grid[i1][j1] == '#') continue;
                int y = i1*m+j1;
                int currMask = mask;
                if (grid[i1][j1] >= 'a' && grid[i1][j1] <= 'z') {
                    currMask |= (1 << m1[grid[i1][j1]]);
                } 
                else if (grid[i1][j1] >= 'A' && grid[i1][j1] <= 'Z') {
                    char key = (char)(grid[i1][j1] + 32);
                    if (!(mask & (1 << m1[key]))) continue;
                }

                if (visited[y][currMask]) continue;
                visited[y][currMask] = 1;
                d[y][currMask] = d[x.first][mask] + 1;
                q.push({y, currMask});
            }
        }

        int reqMask = ((1 << k) - 1);
        int ans = inf;
        for (int i = 0 ; i < n*m ; i++) {
            if (visited[i][reqMask]) ans = min(ans , d[i][reqMask]);
        }

        return (ans == inf ? -1 : ans);
    }
};
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