<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>

struct Node {
    int i;
    int j;
    int step;
    int pickedKeys;
    int state;
    Node(int i, int j, int step, int pickedKeys, int state) {
        this->i = i;
        this->j = j;
        this->step = step;
        this->pickedKeys = pickedKeys;
        this->state = state;
    }
};

class Solution {
public:
    // ["...#.",
    //  "a..@.",
    //  "#..#.",
    //  "b.#B.",
    //  ".##.A"]

    bool isKey(char ch) {
        return ch >= 'a' && ch <= 'z';
    }
    
    bool isLock(char ch) {
        return ch >= 'A' && ch <= 'Z';
    }

    bool keyExists(int pickedKeys, int lockPos) {
        return pickedKeys & (1 << lockPos);
    }

    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size(), n = grid[0].size();
        int totalKeys = 0;
        queue<Node> q;
        unordered_set<int> visited;
        vector<vector<int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if (grid[i][j] == '@') {
                    int curState = (i << 5) | j;
                    q.push(Node(i, j, 0, 0, curState));
                    visited.insert(curState);
                } else if (isKey(grid[i][j])) {
                    totalKeys |= 1 << (grid[i][j] - 'a');
                }
            }
        }

        while (!q.empty()) {
            auto top = q.front();
            q.pop();
            if (top.pickedKeys == totalKeys) {
                return top.step;
            }
            for (auto d : directions) {
                int x = top.i + d[0];
                int y = top.j + d[1];
                int pickedKeys = top.pickedKeys;
                if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] == '#') {
                    continue;
                }
                if (isKey(grid[x][y])) {
                    int pos = grid[x][y] - 'a';
                    pickedKeys |= 1 << pos;
                }
                if (isLock(grid[x][y])) {
                    int pos = grid[x][y] - 'A';
                    if (!keyExists(pickedKeys, pos)) {
                        continue;
                    }
                }
                int curState = (pickedKeys << 10) | (x << 5) | y;
                if (visited.count(curState) == 0) {
                    q.push(Node(x, y, top.step + 1, pickedKeys, curState));
                    visited.insert(curState);
                }
            }
        }
        return -1;
    }
};
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