# 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.

Solution 2: State Compression Dynamic Programming

We can put all the coordinates $(i, j)$ of cells with a value of $0$ into an array $left$. If the value $v$ of a cell is greater than $1$, we put $v-1$ coordinates $(i, j)$ into an array $right$. The problem then becomes that each coordinate $(i, j)$ in $right$ needs to be moved to a coordinate $(x, y)$ in $left$, and we need to find the minimum number of moves.

Let’s denote the length of $left$ as $n$. We can use an $n$-bit binary number to represent whether each coordinate in $left$ is filled by a coordinate in $right$, where $1$ represents being filled, and $0$ represents not being filled. Initially, $f[i] = \infty$, and the rest $f[0]=0$.

Consider $f[i]$, let the number of $1$s in the binary representation of $i$ be $k$. We enumerate $j$ in the range $[0..n)$, if the $j$th bit of $i$ is $1$, then $f[i]$ can be transferred from $f[i \oplus (1 « j)]$, and the cost of the transfer is $cal(left[k-1], right[j])$, where $cal$ represents the Manhattan distance between two coordinates. The final answer is $f[(1 « n) - 1]$.

The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Here, $n$ is the length of $left$, and in this problem, $n \le 9$.

• class Solution {
public int minimumMoves(int[][] grid) {
List<int[]> left = new ArrayList<>();
List<int[]> right = new ArrayList<>();
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (grid[i][j] == 0) {
} else {
for (int k = 1; k < grid[i][j]; ++k) {
}
}
}
}
int n = left.size();
int[] f = new int[1 << n];
Arrays.fill(f, 1 << 30);
f[0] = 0;
for (int i = 1; i < 1 << n; ++i) {
int k = Integer.bitCount(i);
for (int j = 0; j < n; ++j) {
if ((i >> j & 1) == 1) {
f[i] = Math.min(f[i], f[i ^ (1 << j)] + cal(left.get(k - 1), right.get(j)));
}
}
}
return f[(1 << n) - 1];
}

private int cal(int[] a, int[] b) {
return Math.abs(a[0] - b[0]) + Math.abs(a[1] - b[1]);
}
}

• class Solution {
public:
int minimumMoves(vector<vector<int>>& grid) {
using pii = pair<int, int>;
vector<pii> left, right;
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (grid[i][j] == 0) {
left.emplace_back(i, j);
} else {
for (int k = 1; k < grid[i][j]; ++k) {
right.emplace_back(i, j);
}
}
}
}
auto cal = [](pii a, pii b) {
return abs(a.first - b.first) + abs(a.second - b.second);
};
int n = left.size();
int f[1 << n];
memset(f, 0x3f, sizeof(f));
f[0] = 0;
for (int i = 1; i < 1 << n; ++i) {
int k = __builtin_popcount(i);
for (int j = 0; j < n; ++j) {
if (i >> j & 1) {
f[i] = min(f[i], f[i ^ (1 << j)] + cal(left[k - 1], right[j]));
}
}
}
return f[(1 << n) - 1];
}
};

• class Solution:
def minimumMoves(self, grid: List[List[int]]) -> int:
def cal(a: tuple, b: tuple) -> int:
return abs(a[0] - b[0]) + abs(a[1] - b[1])

left, right = [], []
for i in range(3):
for j in range(3):
if grid[i][j] == 0:
left.append((i, j))
else:
for _ in range(grid[i][j] - 1):
right.append((i, j))

n = len(left)
f = [inf] * (1 << n)
f[0] = 0
for i in range(1, 1 << n):
k = i.bit_count()
for j in range(n):
if i >> j & 1:
f[i] = min(f[i], f[i ^ (1 << j)] + cal(left[k - 1], right[j]))
return f[-1]


• func minimumMoves(grid [][]int) int {
left := [][2]int{}
right := [][2]int{}
for i := 0; i < 3; i++ {
for j := 0; j < 3; j++ {
if grid[i][j] == 0 {
left = append(left, [2]int{i, j})
} else {
for k := 1; k < grid[i][j]; k++ {
right = append(right, [2]int{i, j})
}
}
}
}
cal := func(a, b [2]int) int {
return abs(a[0]-b[0]) + abs(a[1]-b[1])
}
n := len(left)
f := make([]int, 1<<n)
f[0] = 0
for i := 1; i < 1<<n; i++ {
f[i] = 1 << 30
k := bits.OnesCount(uint(i))
for j := 0; j < n; j++ {
if i>>j&1 == 1 {
f[i] = min(f[i], f[i^(1<<j)]+cal(left[k-1], right[j]))
}
}
}
return f[(1<<n)-1]
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function minimumMoves(grid: number[][]): number {
const left: number[][] = [];
const right: number[][] = [];
for (let i = 0; i < 3; ++i) {
for (let j = 0; j < 3; ++j) {
if (grid[i][j] === 0) {
left.push([i, j]);
} else {
for (let k = 1; k < grid[i][j]; ++k) {
right.push([i, j]);
}
}
}
}
const cal = (a: number[], b: number[]) => {
return Math.abs(a[0] - b[0]) + Math.abs(a[1] - b[1]);
};
const n = left.length;
const f: number[] = Array(1 << n).fill(1 << 30);
f[0] = 0;
for (let i = 0; i < 1 << n; ++i) {
let k = 0;
for (let j = 0; j < n; ++j) {
if ((i >> j) & 1) {
++k;
}
}
for (let j = 0; j < n; ++j) {
if ((i >> j) & 1) {
f[i] = Math.min(f[i], f[i ^ (1 << j)] + cal(left[k - 1], right[j]));
}
}
}
return f[(1 << n) - 1];
}