Minimum Moves to Spread Stones Over Grid
<p>You are given a <strong>0-indexed</strong> 2D integer matrix <code>grid</code> of size <code>3 * 3</code>, representing the number of stones in each cell. The grid contains exactly <code>9</code> stones, and there can be <strong>multiple</strong> stones in a single cell.</p>

<p>In one move, you can move a single stone from its current cell to any other cell if the two cells share a side.</p>

<p>Return <em>the <strong>minimum number of moves</strong> required to place one stone in each cell</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2850.Minimum%20Moves%20to%20Spread%20Stones%20Over%20Grid/images/example1-3.svg" style="width: 401px; height: 281px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,0],[1,1,1],[1,2,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One possible sequence of moves to place one stone in each cell is: 
1- Move one stone from cell (2,1) to cell (2,2).
2- Move one stone from cell (2,2) to cell (1,2).
3- Move one stone from cell (1,2) to cell (0,2).
In total, it takes 3 moves to place one stone in each cell of the grid.
It can be shown that 3 is the minimum number of moves required to place one stone in each cell.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2850.Minimum%20Moves%20to%20Spread%20Stones%20Over%20Grid/images/example2-2.svg" style="width: 401px; height: 281px;" />
<pre>
<strong>Input:</strong> grid = [[1,3,0],[1,0,0],[1,0,3]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible sequence of moves to place one stone in each cell is:
1- Move one stone from cell (0,1) to cell (0,2).
2- Move one stone from cell (0,1) to cell (1,1).
3- Move one stone from cell (2,2) to cell (1,2).
4- Move one stone from cell (2,2) to cell (2,1).
In total, it takes 4 moves to place one stone in each cell of the grid.
It can be shown that 4 is the minimum number of moves required to place one stone in each cell.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>grid.length == grid[i].length == 3</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 9</code></li>
	<li>Sum of <code>grid</code> is equal to <code>9</code>.</li>
</ul>
Breadth-First Search; Array; Dynamic Programming; Matrix
class Solution:
    def minimumMoves(self, grid: List[List[int]]) -> int:
        q = deque([tuple(tuple(row) for row in grid)])
        vis = set(q)
        ans = 0
        dirs = (-1, 0, 1, 0, -1)
        while 1:
            for _ in range(len(q)):
                cur = q.popleft()
                if all(x for row in cur for x in row):
                    return ans
                for i in range(3):
                    for j in range(3):
                        if cur[i][j] > 1:
                            for a, b in pairwise(dirs):
                                x, y = i + a, j + b
                                if 0 <= x < 3 and 0 <= y < 3 and cur[x][y] < 2:
                                    nxt = [list(row) for row in cur]
                                    nxt[i][j] -= 1
                                    nxt[x][y] += 1
                                    nxt = tuple(tuple(row) for row in nxt)
                                    if nxt not in vis:
                                        vis.add(nxt)
                                        q.append(nxt)
            ans += 1
