LeetCode in Kotlin

2850. Minimum Moves to Spread Stones Over Grid

Medium

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.

Solution

import kotlin.math.abs
import kotlin.math.max
import kotlin.math.min

class Solution {
    fun minimumMoves(grid: Array<IntArray>): Int {
        val a = grid[0][0] - 1
        val b = grid[0][1] - 1
        val c = grid[0][2] - 1
        val d = grid[1][0] - 1
        val f = grid[1][2] - 1
        val g = grid[2][0] - 1
        val h = grid[2][1] - 1
        val i = grid[2][2] - 1
        var minCost = Int.MAX_VALUE
        for (x in min(a, 0)..max(a, 0)) {
            for (y in min(c, 0)..max(c, 0)) {
                for (z in min(i, 0)..max(i, 0)) {
                    for (t in min(g, 0)..max(g, 0)) {
                        val cost: Int =
                            abs(x) + abs(y) + abs(z) + abs(t) + abs((x - a)) + abs(
                                (y - c),
                            ) + abs((z - i)) + abs((t - g)) + abs((x - y + b + c)) + abs(
                                (y - z + i + f),
                            ) + abs((z - t + g + h)) + abs((t - x + a + d))
                        if (cost < minCost) {
                            minCost = cost
                        }
                    }
                }
            }
        }
        return minCost
    }
}

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