LeetCode in Kotlin

1706. Where Will the Ball Fall

Medium

You have a 2-D grid of size m x n representing a box, and you have n balls. The box is open on the top and bottom sides.

Each cell in the box has a diagonal board spanning two corners of the cell that can redirect a ball to the right or to the left.

• A board that redirects the ball to the right spans the top-left corner to the bottom-right corner and is represented in the grid as 1.
• A board that redirects the ball to the left spans the top-right corner to the bottom-left corner and is represented in the grid as -1.

We drop one ball at the top of each column of the box. Each ball can get stuck in the box or fall out of the bottom. A ball gets stuck if it hits a “V” shaped pattern between two boards or if a board redirects the ball into either wall of the box.

Return an array answer of size n where answer[i] is the column that the ball falls out of at the bottom after dropping the ball from the ith column at the top, or -1 _if the ball gets stuck in the box._

Example 1:

Input: grid = [[1,1,1,-1,-1],[1,1,1,-1,-1],[-1,-1,-1,1,1],[1,1,1,1,-1],[-1,-1,-1,-1,-1]]

Output: [1,-1,-1,-1,-1]

Explanation: This example is shown in the photo.

Ball b0 is dropped at column 0 and falls out of the box at column 1.

Ball b1 is dropped at column 1 and will get stuck in the box between column 2 and 3 and row 1.

Ball b2 is dropped at column 2 and will get stuck on the box between column 2 and 3 and row 0.

Ball b3 is dropped at column 3 and will get stuck on the box between column 2 and 3 and row 0.

Ball b4 is dropped at column 4 and will get stuck on the box between column 2 and 3 and row 1.

Example 2:

Input: grid = [[-1]]

Output: [-1]

Explanation: The ball gets stuck against the left wall.

Example 3:

Input: grid = [[1,1,1,1,1,1],[-1,-1,-1,-1,-1,-1],[1,1,1,1,1,1],[-1,-1,-1,-1,-1,-1]]

Output: [0,1,2,3,4,-1]

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• grid[i][j] is 1 or -1.

Solution

class Solution {
    fun findBall(grid: Array<IntArray>): IntArray {
        val m = grid.size
        val n = grid[0].size
        val res = IntArray(n)
        for (j in 0 until n) {
            var currentJ = j
            var currentI = 0
            while (currentJ < n && currentI < m) {
                if (grid[currentI][currentJ] == 1) {
                    currentJ++
                    if (currentJ < n && grid[currentI][currentJ] == 1) {
                        currentI++
                    } else {
                        break
                    }
                } else {
                    currentJ--
                    if (currentJ >= 0 && grid[currentI][currentJ] == -1) {
                        currentI++
                    } else {
                        break
                    }
                }
            }
            if (currentI == m) {
                res[j] = currentJ
            } else {
                res[j] = -1
            }
        }
        return res
    }
}

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