<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 dx[4] = {1,-1,0,0}, dy[4] = {0,0,1,-1};
    int shortestPathAllKeys(vector<string>& grid) {
        int n = grid.size(), m = grid[0].size();

        int keys = 0;
        queue<vector<int>> q;       // {i, j, cost, status}
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '@') q.push({i, j, 0, 0});
                else if(grid[i][j] >= 'a' && grid[i][j] <= 'f') {
                    keys++;
                }
            }
        }
        int finalStatus = (1 << keys) - 1;
        vector<vector<vector<int>>> visited(n, vector<vector<int>>(m, vector<int>(finalStatus+1, 0)));

        while(!q.empty()) {
            auto temp = q.front();
            q.pop();

            int x = temp[0], y = temp[1], cost = temp[2], status = temp[3];
            if(status == finalStatus) {
                return cost;
            }

            for(int k = 0; k < 4; k++) {
                int nx = x+dx[k], ny = y+dy[k];

                if(nx < 0 || ny < 0 || nx >= n || ny >= m || grid[nx][ny] == '#') {
                    continue;
                }
                char c = grid[nx][ny];
                if(c >= 'a' && c <= 'f') {
                    int new_status = status | (1 << (c-'a'));
                    if(visited[nx][ny][new_status] == 0) {
                        visited[nx][ny][new_status] = 1;
                        q.push({nx, ny, cost+1, new_status});
                    }
                }
                else if(c >= 'A' && c <= 'F') {
                    if(((status >> (c-'A')) & 1) == 1) {
                        if (!visited[nx][ny][status]) {
                            visited[nx][ny][status] = 1;
                            q.push({nx, ny, cost + 1, status});
                        }
                    }
                }
                else {
                    if(visited[nx][ny][status] == 0) {
                        visited[nx][ny][status] = 1;
                        q.push({nx, ny, cost+1, status});
                    }
                }



            }
        }
        return -1;
    }
};
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