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

        int x, y, len=0;

        set<vector<int>> s;
        queue<vector<int>> q;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                char c = a[i][j];
                if (c == '@') {
                    x = i;
                    y = j;
                    a[i][j] = '.';
                }
                if (c >= 'a' && c <= 'f') {
                    len = max(len, c - 'a' + 1);
                }
            }
        }
        
        s.insert({x, y, 0});
        q.push({x, y, 0});

        int ans = -1;
        while (q.size() != 0) {
            int sz = q.size();
            ++ans;
            while (sz--) {
                int i = q.front()[0];
                int j = q.front()[1];
                int keys = q.front()[2];
                q.pop();

                // Do I have all the keys?
                if (keys == ((1 << len) - 1))
                    return ans;

                for (int k = 0; k < 4; k++) {
                    int kk = keys;
                    int ni = i + r[k];
                    int nj = j + c[k];

                    if (ni < 0 || nj < 0 || ni >= m || nj >= n ||
                        a[ni][nj] == '#')
                        continue;
                    char c = a[ni][nj];


                    if (c <= 'z' && c >= 'a') {
                        // Setting the bit where key found
                        kk = kk | (1 << (c - 'a'));
                        if (kk == ((1 << len) - 1))
                            return ans + 1;
                    } 

                    if (c <= 'Z' && c >= 'A') {
                        if ((kk & (1 << (c - 'A'))) == 0)
                            continue;
                    }

                    // Check if cell vis or not with same set of keys
                    if (s.find({ni, nj, kk}) != s.end())
                        continue;
                    s.insert({ni, nj, kk});
                    q.push({ni, nj, kk});
                }
            }
        }
        return -1;
    }
};
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