<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:
    
    vector<vector<int>> directions{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    
    int shortestPathAllKeys(vector<string>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        
        queue<vector<int>> que; //{x, y, steps, current_key_status_decimal}
        int count = 0;
        for(int i = 0; i<m; i++) {
            
            for(int j = 0; j<n; j++) {
                
                if(grid[i][j] == '@') {
                    que.push({i, j, 0, 0});
                } else if(grid[i][j] >= 'a' && grid[i][j] <= 'f') {
                    count++;
                }
                
            }
            
        }
        
        /*
            If we have 3 keys - '111'  - Decimal value = 8 = 2^3 - 1
            If we have 4 keys - '1111' - Decimal value = 15 = 2^4 - 1
        */
        
        int final_key_status_decimal = pow(2, count) - 1;
        
        
        int visited[m][n][final_key_status_decimal+1];
        memset(visited, 0, sizeof(visited));
        
        while(!que.empty()) {
            
            auto temp = que.front();
            que.pop();
            
            int i                            = temp[0];
            int j                            = temp[1];
            int steps                        = temp[2];
            int current_key_status_decimal   = temp[3];
            
            if(current_key_status_decimal == final_key_status_decimal) {
                return steps;
            }
            
            for(vector<int> & dir : directions) {
                
                int new_i = i + dir[0];
                int new_j = j + dir[1];
                
                if(new_i >= 0 && new_i < m && new_j >= 0 && new_j < n && grid[new_i][new_j] != '#') {
                    char ch = grid[new_i][new_j];
                        
                    if(grid[new_i][new_j] >= 'A' && grid[new_i][new_j] <= 'F') { //Lock
                        
                        if(visited[new_i][new_j][current_key_status_decimal] == 0 && 
                          ((current_key_status_decimal >> (ch-'A')) & 1) == 1) { //Already have that key then we unlock
                            visited[new_i][new_j][current_key_status_decimal] = 1;
                            que.push({new_i, new_j, steps+1, current_key_status_decimal});
                        }
                        
                    } else if(grid[new_i][new_j] >= 'a' && grid[new_i][new_j] <= 'f') { //Key
                        
                        int newKey_status_decimal = current_key_status_decimal | (1 << (ch - 'a'));
                        
                        if(visited[new_i][new_j][newKey_status_decimal] == 0) {
                            visited[new_i][new_j][newKey_status_decimal] = 1;
                            que.push({new_i, new_j, steps+1, newKey_status_decimal});
                        }
                        
                    } else {
                        if(visited[new_i][new_j][current_key_status_decimal] == 0) {
                            visited[new_i][new_j][current_key_status_decimal] = 1;
                            que.push({new_i, new_j, steps+1, current_key_status_decimal});
                        }
                    }
                    
                }
                
            }
            
        }
        
        return -1;
    }
};
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