#2850

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Moves to Spread Stones Over Grid

Solve the problem of distributing 9 stones across a 3x3 grid with minimal moves using state transition dynamic programming.

array dynamic programming breadth first search matrix

Problem Statement

You are given a 3x3 grid where each cell contains some number of stones. The total number of stones in the grid is exactly 9, and there can be multiple stones in a single cell.

In one move, you can move a single stone from one cell to an adjacent one (horizontally or vertically). Your task is to return the minimum number of moves needed to place one stone in each cell.

Examples

Example 1

Input: grid = [[1,1,0],[1,1,1],[1,2,1]]

Output: 3

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

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.

Solution Approach

State Transition Dynamic Programming

This problem can be solved using state transition dynamic programming, where we track the movement of stones from cells with excess stones to those with fewer. By evaluating different configurations and transitions, we can compute the minimum number of moves.

Breadth-First Search

Breadth-First Search (BFS) is useful for exploring the grid and determining the shortest path for moving stones from one cell to another. This ensures we are making optimal moves at each step.

Grid Transformation Strategy

We perform grid transformation by shifting stones from crowded cells to those with fewer stones, ensuring that every cell eventually holds exactly one stone. Tracking these transformations leads to the optimal solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the final algorithm used. If BFS is used, the complexity would be proportional to the number of possible states, which is influenced by grid size and stone distribution. State transition dynamic programming also impacts the complexity, potentially increasing with the number of configurations to consider.

What Interviewers Usually Probe

Evaluate the candidate's understanding of state transition dynamic programming.
Assess the candidate's ability to efficiently perform grid transformations.
Look for a clear explanation of BFS and how it ensures minimal move calculations.

Common Pitfalls or Variants

Common pitfalls

Forgetting to track the optimal state transitions, leading to unnecessary moves.
Overlooking edge cases where certain cells may need more or fewer stones than others.
Misunderstanding the grid's adjacency rules, leading to incorrect stone placement.

Follow-up variants

What if the grid size increases to 4x4? The problem's state transitions will expand, requiring more complex dynamic programming.
What if there are more than 9 stones? The challenge will include distributing the stones more efficiently across larger grids.
What if stones are placed randomly in the grid? A solution must account for irregular stone distributions and still aim for the minimum move count.

FAQ

What is the primary pattern used to solve Minimum Moves to Spread Stones Over Grid?

The primary pattern is state transition dynamic programming, which helps optimize the number of moves by tracking stone distribution.

How do I apply dynamic programming to this problem?

Dynamic programming helps track the movement of stones from one cell to another, minimizing the total number of moves needed to achieve the goal state.

What role does BFS play in this problem?

BFS is used to explore the grid and find the shortest path for moving stones, ensuring the solution is efficient and optimal.

How do I handle edge cases in this problem?

Edge cases include ensuring each cell eventually holds exactly one stone, while handling possible irregular distributions efficiently.

Can GhostInterview assist with this type of state transition problem?

Yes, GhostInterview provides insights and strategies to solve state transition problems and helps guide your approach towards the optimal solution.

terminal

Solution

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.

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

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.

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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