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
function minimumMoves(grid: number[][]): number {
    const q: string[] = [f(grid)];
    const vis: Set<string> = new Set([f(grid)]);
    const dirs: number[] = [-1, 0, 1, 0, -1];

    for (let ans = 0; ; ans++) {
        let sz = q.length;
        while (sz-- > 0) {
            const p = q.shift()!;
            if (p === '111111111') {
                return ans;
            }
            const cur = g(p);

            for (let i = 0; i < 3; i++) {
                for (let j = 0; j < 3; j++) {
                    if (cur[i][j] > 1) {
                        for (let d = 0; d < 4; d++) {
                            const x = i + dirs[d],
                                y = j + dirs[d + 1];
                            if (x >= 0 && x < 3 && y >= 0 && y < 3 && cur[x][y] < 2) {
                                const nxt = cur.map(row => [...row]);
                                nxt[i][j]--;
                                nxt[x][y]++;
                                const s = f(nxt);
                                if (!vis.has(s)) {
                                    vis.add(s);
                                    q.push(s);
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

function f(grid: number[][]): string {
    return grid.flat().join('');
}

function g(s: string): number[][] {
    return Array.from({ length: 3 }, (_, i) =>
        Array.from({ length: 3 }, (_, j) => Number(s[i * 3 + j])),
    );
}
