<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) {
        int m = grid.size(), n = grid[0].size();
        int startX = -1, startY = -1;
        string allKeys = "";

        // Find the starting point and determine all keys
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '@') {
                    startX = i;
                    startY = j;
                } else if (islower(grid[i][j])) {
                    allKeys += grid[i][j];
                }
            }
        }

        queue<tuple<int, int, string, int>> q;  // (x, y, collectedKeys, steps)
        set<tuple<int, int, string>> visited;

        q.push({startX, startY, "", 0});
        visited.insert({startX, startY, ""});

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

        while (!q.empty()) {
            auto [x, y, keys, steps] = q.front();
            q.pop();

            // Check if all keys have been collected
            if (keys.size() == allKeys.size()) {
                return steps;
            }

            // Explore in four directions
            for (auto& dir : directions) {
                int nx = x + dir.first;
                int ny = y + dir.second;
                if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] != '#') {
                    char cell = grid[nx][ny];
                    string nextKeys = keys;

                    if (islower(cell) && nextKeys.find(cell) == string::npos) {
                        nextKeys += cell;
                        // sort(nextKeys.begin(), nextKeys.end());  // Keep keys sorted to avoid duplicates in different orders
                    }

                    if (isupper(cell) && nextKeys.find(tolower(cell)) == string::npos) {
                        continue;  // Cannot pass lock without key
                    }

                    if (visited.find({nx, ny, nextKeys}) == visited.end()) {
                        q.push({nx, ny, nextKeys, steps + 1});
                        visited.insert({nx, ny, nextKeys});
                    }
                }
            }
        }

        return -1;  // If it's impossible to collect all keys
    }
};

// [
// "@ . a . ."
// "# # # . #"a
// "b . A . B"
// ]
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