Introduction

In this blog post, we will explore how to solve the LeetCode problem "2850" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the approach to solve this problem:

• Model as a Bipartite Graph: Think of the grid cells with excess stones as "sources" and the cells needing stones as "sinks." The cost of moving a stone between two cells is the Manhattan distance between them.
• Minimum Cost Maximum Flow: The problem now becomes finding the minimum cost to transport stones from sources to sinks, which can be solved using a min-cost max-flow algorithm. Since the total excess and deficiency are equal (both sum to the same value, ensuring max flow is trivially the total number of stones that need moving), we only need a min-cost flow algorithm.

• Simplified Solution for 3x3 Grid: For the specific 3x3 grid case, a simplified approach focusing on iterating through the excess and deficiency cells can yield the minimum cost more efficiently than a full min-cost max-flow implementation.

• Runtime Complexity: O(N^2), where N is the number of cells (9 in this case, so practically O(1)). Storage Complexity: O(1).

Code

    def minimum_moves(grid):
    """
    Calculates the minimum number of moves required to place one stone in each cell of the grid.

    Args:
        grid: A 3x3 2D list of integers representing the number of stones in each cell.

    Returns:
        The minimum number of moves required.
    """

    excess = []
    deficient = []
    for i in range(3):
        for j in range(3):
            if grid[i][j] > 1:
                excess.extend([(i, j)] * (grid[i][j] - 1))
            elif grid[i][j] < 1:
                deficient.extend([(i, j)] * (1 - grid[i][j]))

    moves = 0
    i = 0
    j = 0
    while i < len(excess) and j < len(deficient):
        x1, y1 = excess[i]
        x2, y2 = deficient[j]
        moves += abs(x1 - x2) + abs(y1 - y2)
        i += 1
        j += 1

    return moves