<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 State {
    int row;
    int col;
    int keys;
    State(int row, int col): row(row), col(col), keys(0) {};
    State(int row, int col, int keys): row(row), col(col), keys(keys) {};
};
class Solution {
public:
    vector<pair<int,int>> directions = {{1,0},{0,1},{-1,0},{0,-1}};
    int shortestPathAllKeys(vector<string>& grid) {
        int row_size = grid.size();
        int col_size = grid[0].length();

        queue<State*> q;

        int all_keys_sequence = 0;
        int start_i = -1;
        int start_j = -1;
        for(int i = 0; i < row_size; ++i) {
            for(int j = 0; j < col_size; ++j) {
                if(grid[i][j] == '@') {
                    start_i = i;
                    start_j = j;
                }
                if(grid[i][j] >= 'a' && grid[i][j] <= 'f') {
                    all_keys_sequence |= (1 << (grid[i][j] - 'a'));
                }
            }
        }
        vector<vector<vector<bool>>> visited(row_size, vector<vector<bool>>(col_size, vector<bool>(64, false)));

        q.push(new State(start_i, start_j));
        int steps = 0;
        while(!q.empty()) {
            int q_size = q.size();
            for(int i = 0; i < q_size; ++i) {
                State* cur_state = q.front();
                q.pop();
                int& cur_row = cur_state->row;
                int& cur_col = cur_state->col;
                int& key_state = cur_state->keys;

                for(pair<int,int>& direction : directions) {
                    int new_row = cur_row + direction.first;
                    int new_col = cur_col + direction.second;
                    if(new_row < 0 || new_col < 0 || new_row >= row_size || new_col >= col_size ||
                            grid[new_row][new_col] == '#') continue;

                    char& grid_char = grid[new_row][new_col];
                    if(grid_char >= 'A' && grid_char <= 'F') {
                        if((key_state & (1 << (grid_char - 'A'))) == 0) {
                            continue;
                        }
                    }

                    int new_key_state = key_state;

                    if(grid_char >= 'a' && grid_char <= 'f') {
                        new_key_state |= (1 << (grid_char - 'a'));

                        if(new_key_state == all_keys_sequence) {
                            return steps + 1;
                        }
                    }
                    if(visited[new_row][new_col][new_key_state]) continue;
                    visited[new_row][new_col][new_key_state] = true;
                    q.push(new State(new_row, new_col, new_key_state));
                }


            }
            ++steps;
        }

        return -1;

    }
};
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