1706. Where Will the Ball Fall

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

The straightforward idea is to go through the top of matrix for grid[0][i] where 0 <= i <= n, by DFS and return the results of each entry. Instead of filtering the cases that the ball got stuck, we only check the case that the ball can pass to the next level.

Once it reached the (last level + 1) of grids, the col value is the answer we want.

int m = 0;
int n = 0;
public int[] findBall(int[][] grid) {
    m = grid.length;
    n = grid[0].length;
    int[] ret = new int[n];
    for(int i = 0; i < n; i++) {
        ret[i] = dfs(0, i, grid);
    }
    return ret;
}

public int dfs(int i, int j, int[][] grid) {

    if(i == m) return j;
    if(grid[i][j] == 1 && j+1 < n && grid[i][j+1] == 1){
        return dfs(i+1, j+1, grid);
    }
    if(grid[i][j] == -1 && j-1 >= 0 && grid[i][j-1] == -1){
        return dfs(i+1, j-1, grid);
    }
    return -1;
}