<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 count1s (int num){
        int count = 0;
        for(int i = 0; i < 7; i++){
            if(((num >> i) & 1)  == 1) count++;
        }
        return count;
    }

    int shortestPathAllKeys(vector<string>& grid) {
        int r = grid.size(), c = grid[0].size();

        // index is key , 0 means not visited, 1 means visited
        vector<vector<vector<int>>> visited(r, vector<vector<int>>(c, vector<int>(65, 0)));

        // finding starting point
        int sx, sy;
        int total_keys = 0;

        for(int i = 0; i < r; i++){
            for(int j = 0; j < c; j++){
                if(grid[i][j] == '@'){
                    sx = i;
                    sy = j;
                }
                else if(grid[i][j] >= 'a' && grid[i][j] <= 'f') total_keys++;
            }
        }

        queue<vector<int>> q;

        q.push({sx, sy, 0, 0});
        visited[sx][sy][0] = 1;

        vector<vector<int>> dirxns = {{-1,0}, {1, 0}, {0, -1}, {0, 1}};

        while(!q.empty()){
            auto temp = q.front();
            q.pop();
            int x = temp[0], y = temp[1], dist = temp[2], key = temp[3];

            if(count1s(key) == total_keys) return dist;

            for(int i = 0; i < 4; i++){

                int xnbr = x + dirxns[i][0], ynbr = y + dirxns[i][1];
                int keynbr = key;

                // outside the boundary
                if(xnbr < 0 || xnbr >= r || ynbr < 0 || ynbr >= c) continue;
                // wall
                if(grid[xnbr][ynbr] == '#') continue;
                
                // lock
                if(grid[xnbr][ynbr] >= 'A' && grid[xnbr][ynbr] <= 'F'){

                    int lock = grid[xnbr][ynbr] - 'A';
                    // dont have key
                    if(((key>>lock)&1) == 0) continue;
                }

                // already visited
                if(visited[xnbr][ynbr][key] == 1) continue;

                // push into the queue

                // key updation
                if(grid[xnbr][ynbr] >= 'a' && grid[xnbr][ynbr] <= 'f'){
                    int i = grid[xnbr][ynbr] - 'a';

                    keynbr = (key | (1 << i));
                }

                q.push({xnbr, ynbr, dist+1, keynbr});

                visited[xnbr][ynbr][keynbr] = 1;

            }

        }

        return -1;
    }
};
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