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

          vector<vector<vector<bool>>>vis(grid.size(),vector<vector<bool>>(grid[0].size(),vector<bool>(64,false)));

          int keys = 0;
          int start_row , start_col;
          for(int i = 0 ; i < grid.size() ; i++){
             for(int j = 0 ; j < grid[i].length() ; j++){
                 if(grid[i][j] == '@'){
                     start_row = i;
                     start_col = j;
                 }
                 else if(grid[i][j] >= 'a' and grid[i][j] <= 'z'){ //check
                    keys++;
                 }
             }
          }
          
          queue<vector<int>>q;
          q.push({start_row,start_col,0,0});
          vis[start_row][start_col][0] = true;

          while(q.size() > 0)
          {
               int row = q.front()[0];
               int col = q.front()[1];
               int mask = q.front()[2];
               int steps = q.front()[3];

               q.pop();

               int req_ans = pow(2,keys);
               if(mask == (req_ans-1)) return steps;

               for(int i = 0 ; i < dir.size() ; i++){
                   int r = dir[i][0]; int c = dir[i][1];
                   if(isValid(row+r,col+c) == true)
                   {
                        if(grid[row+r][col+c] == '.'){
                             if(vis[row+r][col+c][mask] == false){
                                  vis[row+r][col+c][mask] = true;
                                  q.push({row+r,col+c,mask,steps+1});    //with the curr mask we are trying to get on the cell row+r , col+c
                             }
                        }
                        else if(grid[row+r][col+c] == '#'){
                             continue;
                        }
                        else if(grid[row+r][col+c] == '@'){
                             if(vis[row+r][col+c][mask] == false){
                                  vis[row+r][col+c][mask] = true;
                                  q.push({row+r,col+c,mask,steps+1});    //with the curr mask we are trying to get on the cell row+r , col+c
                             }
                        }
                        else if(grid[row+r][col+c] >= 'a' and grid[row+r][col+c] <= 'z'){
                             if(vis[row+r][col+c][mask] == false){
                                  vis[row+r][col+c][mask] = true;
                                  int updated_mask = mask | (1 << (grid[row+r][col+c] - 'a'));
                                  q.push({row+r,col+c,updated_mask,steps+1});    //with the curr mask we are trying to get on the cell row+r , col+c
                             }
                        }
                        else{
                             if(vis[row+r][col+c][mask] == false){
                                 char lock = (grid[row+r][col+c]);
                                 char key = tolower(lock);
                                 //checking if in the curr mask we have the key for this lock or not
                                 int found = (mask & (1 << (key - 'a'))); 
                                 if(found > 0) q.push({row+r,col+c,mask,steps+1});   //finally we can land on the cell row+r col+c after unlocking it

                                 vis[row+r][col+c][mask] = true;
                             } 
                        } 
                   }
               }
          }
          return -1;
    }
    bool isValid(int row, int col){
         if(row >=0 and row < n and col >=0 and col < m) return true;
         return false;
    }
};
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