<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) {
        set<string> states;
        queue<vector<int>> q;
        int n = grid.size();
        int m = grid[0].size();
        int allKeys = 0;
        
        // Find initial state and push into queue
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '@') {
                    q.push({i, j, 0, 0});  // {x, y, keys_collected, steps}
                    states.insert(to_string(i) + "," + to_string(j) + ",0");
                } else if(islower(grid[i][j])) {
                    allKeys |= (1 << (grid[i][j] - 'a'));
                }
            }
        }
        
        vector<vector<int>> moves = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
        
        // Start BFS
        while(!q.empty()) {
            auto state = q.front();
            q.pop();
            
            int x = state[0], y = state[1], keys = state[2], steps = state[3];
            
            // Check if all keys are collected
            if(keys == allKeys) return steps;
            
            for(auto& move : moves) {
                int nextx = x + move[0];
                int nexty = y + move[1];
                
                if(nextx >= 0 && nexty >= 0 && nextx < n && nexty < m) {
                    char c = grid[nextx][nexty];
                    int newKeys = keys;
                    
                    if(c == '#') continue;  // Wall
                    if(islower(c)) {
                        newKeys |= 1 << (c - 'a');  // Collect key
                    } else if(isupper(c)) {
                        if((keys & (1 << (c - 'A'))) == 0) continue;  // Door without key
                    }
                    
                    string newState = to_string(nextx) + "," + to_string(nexty) + "," + to_string(newKeys);
                    
                    if(!states.count(newState)) {
                        q.push({nextx, nexty, newKeys, steps + 1});
                        states.insert(newState);
                    }
                }
            }
        }
        
        return -1;  // If all keys cannot be collected
    }
};

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