<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 shortestPathAllKeys(vector<string>& grid) {

        //crux is we can revisit same node provided we have diff key comb
        // like (1,1) we vis with key="a" now in future (1,1) again with "abc" is valid
        int key_count = 0;
        int n = grid.size();
        int m = grid[0].size();
        queue<pair<pair<pair<int,int>,int>,string>> Q; // (i,j),steps,keys
        int st_i = -1, st_j = -1;

        // Identify starting point and count the total number of keys
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] >= 'a' && grid[i][j] <= 'z') key_count++;
                if(grid[i][j] == '@') {
                    st_i = i;
                    st_j = j;
                }
            }
        }

        Q.push({{{st_i, st_j}, 0}, ""});
        set<pair<pair<int,int>, string>> St; // ((i,j),curr_all_key_holding)
        St.insert({{st_i, st_j}, ""});

        // Directions for movement: down, up, right, left
        vector<pair<int, int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

        while(!Q.empty()) {
            auto it = Q.front();
            int i = it.first.first.first;
            int j = it.first.first.second;
            int steps = it.first.second;
            string keys = it.second;
            Q.pop();

            // If we've collected all keys, return the number of steps
            if(keys.size() == key_count) return steps;

            for(auto dir : directions) {
                int ni = i + dir.first; //new row
                int nj = j + dir.second; //new col

                if(ni >= 0 && ni < n && nj >= 0 && nj < m && grid[ni][nj] != '#') {
                    char cell = grid[ni][nj]; //new cell 
                    string new_keys = keys;

                    // If it's a key, add it to the keys string if not already present
                    if(cell >= 'a' && cell <= 'z' && keys.find(cell) == string::npos) {
                        new_keys += cell;
                    }

                    // If it's a lock, check if we have the key
                    if(cell >= 'A' && cell <= 'Z' && keys.find(cell + 32) == string::npos) {
                        continue; // Skip this cell as we don't have the key
                    }

                    // Only proceed if the state hasn't been visited with the same keys
                    if(St.find({{ni, nj}, new_keys}) == St.end()) {
                        St.insert({{ni, nj}, new_keys});
                        Q.push({{{ni, nj}, steps + 1}, new_keys});
                    }
                }
            }
        }

        return -1;
    }
};

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