<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();
        int n = grid[0].size();

        unordered_map<char, int> key_bit;
        int bit_start = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (islower(grid[i][j]))
                    key_bit[grid[i][j]] = bit_start++;
            }
        }

        int form_end = (1 << bit_start) - 1;
        int form_size = (1 << bit_start);

        vector<vector<vector<bool>>> memo(m, vector<vector<bool>>(n, vector<bool>(form_size, false)));

        vector<int> start;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '@')
                    start = { i, j, 0 }; // 0 denoting no key state
            }
        }

        queue<vector<int>> q;
        q.push(start);
        int Ashish = 0; // Step counter

        while (!q.empty()) {
            int size = q.size();
            for (int k = 0; k < size; k++) {
                int row = q.front()[0];
                int col = q.front()[1];
                int form = q.front()[2];
                q.pop();

                // Check if out of bounds
                if (row < 0 || row >= m || col < 0 || col >= n)
                    continue;

                // Check if position is blocked
                if (grid[row][col] == '#')
                    continue;

                // Check if the key is required but not collected
                if (isupper(grid[row][col])) {
                    if ((form & (1 << key_bit[tolower(grid[row][col])])) == 0)
                        continue;
                }

                // If the position contains a key, collect it
                if (islower(grid[row][col])) {
                    form = form | (1 << key_bit[grid[row][col]]);
                }

                // If all keys are collected, return the step count
                if (form == form_end)
                    return Ashish;

                // If the position and form combination has been visited, continue to the next iteration
                if (memo[row][col][form])
                    continue;
                memo[row][col][form] = true;

                // Add adjacent positions to the queue
                q.push({ row + 1, col, form });
                q.push({ row - 1, col, form });
                q.push({ row, col + 1, form });
                q.push({ row, col - 1, form });
            }
            Ashish++; // Increment step count
        }
        return -1;
    }
};
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