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2850. Minimum Moves to Spread Stones Over Grid

Description

You are given a 0-indexed 2D integer matrix grid of size 3 * 3, representing the number of stones in each cell. The grid contains exactly 9 stones, and there can be multiple stones in a single cell.

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

Return the minimum number of moves required to place one stone in each cell.

 

Example 1:

Input: grid = [[1,1,0],[1,1,1],[1,2,1]]
Output: 3
Explanation: 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.

Example 2:

Input: grid = [[1,3,0],[1,0,0],[1,0,3]]
Output: 4
Explanation: 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.

 

Constraints:

  • grid.length == grid[i].length == 3
  • 0 <= grid[i][j] <= 9
  • Sum of grid is equal to 9.

Solutions

Solution 1: Naive BFS

The problem is essentially finding the shortest path from the initial state to the target state in a state graph, so we can use BFS to solve it. The initial state is grid, and the target state is [[1, 1, 1], [1, 1, 1], [1, 1, 1]]. In each operation, we can move a stone greater than $1$ from a cell to an adjacent cell that does not exceed $1$. If the target state is found, we can return the current layer number, which is the minimum number of moves.

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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

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class Solution {
    public int minimumMoves(int[][] grid) {
        Deque<String> q = new ArrayDeque<>();
        q.add(f(grid));
        Set<String> vis = new HashSet<>();
        vis.add(f(grid));
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int ans = 0;; ++ans) {
            for (int k = q.size(); k > 0; --k) {
                String p = q.poll();
                if ("111111111".equals(p)) {
                    return ans;
                }
                int[][] cur = g(p);
                for (int i = 0; i < 3; ++i) {
                    for (int j = 0; j < 3; ++j) {
                        if (cur[i][j] > 1) {
                            for (int d = 0; d < 4; ++d) {
                                int x = i + dirs[d];
                                int y = j + dirs[d + 1];
                                if (x >= 0 && x < 3 && y >= 0 && y < 3 && cur[x][y] < 2) {
                                    int[][] nxt = new int[3][3];
                                    for (int r = 0; r < 3; ++r) {
                                        for (int c = 0; c < 3; ++c) {
                                            nxt[r][c] = cur[r][c];
                                        }
                                    }
                                    nxt[i][j]--;
                                    nxt[x][y]++;
                                    Strin