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.lengthn == grid[i].length1 <= m, n <= 100grid[i][j]is1or-1.
class Solution:
def findBall(self, grid: List[List[int]]) -> List[int]:
m, n = len(grid), len(grid[0])
def dfs(i, j):
nonlocal m, n
if i == m:
return j
if j == 0 and grid[i][j] == -1:
return -1
if j == n - 1 and grid[i][j] == 1:
return -1
if grid[i][j] == 1 and grid[i][j + 1] == -1:
return -1
if grid[i][j] == -1 and grid[i][j - 1] == 1:
return -1
return dfs(i + 1, j + 1) if grid[i][j] == 1 else dfs(i + 1, j - 1)
return [dfs(0, j) for j in range(n)]class Solution {
private int m;
private int n;
private int[][] grid;
public int[] findBall(int[][] grid) {
m = grid.length;
n = grid[0].length;
this.grid = grid;
int[] ans = new int[n];
for (int j = 0; j < n; ++j) {
ans[j] = dfs(0, j);
}
return ans;
}
private int dfs(int i, int j) {
if (i == m) {
return j;
}
if (j == 0 && grid[i][j] == -1) {
return -1;
}
if (j == n - 1 && grid[i][j] == 1) {
return -1;
}
if (grid[i][j] == 1 && grid[i][j + 1] == -1) {
return -1;
}
if (grid[i][j] == -1 && grid[i][j - 1] == 1) {
return -1;
}
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);
}
}class Solution {
public:
int m, n;
vector<vector<int>> grid;
vector<int> findBall(vector<vector<int>>& grid) {
this->grid = grid;
m = grid.size();
n = grid[0].size();
vector<int> ans(n);
for (int j = 0; j < n; ++j) ans[j] = dfs(0, j);
return ans;
}
int dfs(int i, int j) {
if (i == m) return j;
if (j == 0 && grid[i][j] == -1) return -1;
if (j == n - 1 && grid[i][j] == 1) return -1;
if (grid[i][j] == 1 && grid[i][j + 1] == -1) return -1;
if (grid[i][j] == -1 && grid[i][j - 1] == 1) return -1;
return grid[i][j] == 1 ? dfs(i + 1, j + 1) : dfs(i + 1, j - 1);
}
};func findBall(grid [][]int) []int {
m, n := len(grid), len(grid[0])
var dfs func(i, j int) int
dfs = func(i, j int) int {
if i == m {
return j
}
if j == 0 && grid[i][j] == -1 {
return -1
}
if j == n-1 && grid[i][j] == 1 {
return -1
}
if grid[i][j] == 1 && grid[i][j+1] == -1 {
return -1
}
if grid[i][j] == -1 && grid[i][j-1] == 1 {
return -1
}
if grid[i][j] == 1 {
return dfs(i+1, j+1)
}
return dfs(i+1, j-1)
}
var ans []int
for j := 0; j < n; j++ {
ans = append(ans, dfs(0, j))
}
return ans
}function findBall(grid: number[][]): number[] {
const m = grid.length;
const n = grid[0].length;
const res = new Array(n).fill(0);
const dfs = (i: number, j: number) => {
if (i === m) {
return j;
}
if (grid[i][j] === 1) {
if (j === n - 1 || grid[i][j + 1] === -1) {
return -1;
}
return dfs(i + 1, j + 1);
} else {
if (j === 0 || grid[i][j - 1] === 1) {
return -1;
}
return dfs(i + 1, j - 1);
}
};
for (let i = 0; i < n; i++) {
res[i] = dfs(0, i);
}
return res;
}impl Solution {
fn dfs(grid: &Vec<Vec<i32>>, i: usize, j: usize) -> i32 {
if i == grid.len() {
return j as i32;
}
if grid[i][j] == 1 {
if j == grid[0].len() - 1 || grid[i][j + 1] == -1 {
return -1;
}
Self::dfs(grid, i + 1, j + 1)
} else {
if j == 0 || grid[i][j - 1] == 1 {
return -1;
}
Self::dfs(grid, i + 1, j - 1)
}
}
pub fn find_ball(grid: Vec<Vec<i32>>) -> Vec<i32> {
let m = grid.len();
let n = grid[0].len();
let mut res = vec![0; n];
for i in 0..n {
res[i] = Self::dfs(&grid, 0, i);
}
res
}
}