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Minimum Moves to Spread Stones Over Grid

Difficulty: Medium

Problem 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.

Key Insights

  • The grid is fixed at a size of 3x3, making it feasible to consider all possible configurations and movements.
  • The goal is to ensure that each cell contains exactly one stone.
  • The problem can be approached using a Breadth-First Search (BFS) to explore all possible moves efficiently.
  • The Manhattan distance can be utilized to calculate the cost of moving stones from one cell to another.

Space and Time Complexity

Time Complexity: O(1) - The maximum number of states is limited due to the fixed size of the grid, allowing for a constant time complexity in a practical sense. However, in a more general case, it could be considered O(N) where N is the number of stones. Space Complexity: O(1) - The space used for the queue and visited states is constant because the grid size does not change.

Solution

To solve the problem, we can use a Breadth-First Search (BFS) algorithm. We will start from the initial configuration of stones and explore all possible moves until we reach the desired configuration where each cell contains exactly one stone.

  1. Count the number of stones in each cell.
  2. Use BFS to explore all possible configurations by moving stones from one cell to an adjacent cell.
  3. Keep track of the number of moves taken to reach each configuration.
  4. Once we reach a configuration where each cell has exactly one stone, return the number of moves.

The BFS will ensure that we explore the shortest path (minimum moves) to the solution.

Code Solutions

from collections import deque

def minMovesToSpreadStones(grid):
    target = (1, 1, 1, 1, 1, 1, 1, 1, 1)  # Desired state
    initial = [grid[i][j] for i in range(3) for j in range(3)]
    
    def bfs(initial):
        queue = deque([(initial, 0)])  # (current state, moves)
        visited = set()
        visited.add(tuple(initial))
        
        while queue:
            current, moves = queue.popleft()

            if tuple(current) == target:
                return moves
            
            # Generate next states
            for i in range(3):
                for j in range(3):
                    if current[i * 3 + j] > 1:  # Can move from this cell
                        for di, dj in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                            ni, nj = i + di, j + dj
                            if 0 <= ni < 3 and 0 <= nj < 3:  # Stay within bounds
                                next_state = current[:]
                                next_state[i * 3 + j] -= 1
                                next_state[ni * 3 + nj] += 1
                                state_tuple = tuple(next_state)
                                if state_tuple not in visited:
                                    visited.add(state_tuple)
                                    queue.append((next_state, moves + 1))
        
        return -1  # Should not reach here for valid input
    
    return bfs(initial)

# Example usage
print(minMovesToSpreadStones([[1,1,0],[1,1,1],[1,2,1]]))  # Output: 3
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