Formatted question description: https://leetcode.ca/all/1706.html
1706. Where Will the Ball Fall
Level
Medium
Description
You have a 2D 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 topleft corner to the bottomright corner and is represented in the grid as
1
.  A board that redirects the ball to the left spans the topright corner to the bottomleft 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.
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 100
grid[i][j]
is1
or1
.
Solution
For each ball, use the corresponding elements in grid
to determine whether the ball moves to the right cell or the left cell. If the ball moves out of the bound, which means the board redirects the ball into either wall, return 1. If the two adjacent cells have different values in grid
, which means the ball hits a “V” shaped pattern between two boards, return 1. If neither case happens, the ball moves to the down cell. Repeat the process until the ball gets stucked or the ball reaches the bottom, and then the corresponding answer for the ball can be obtained.
Java

class Solution { public int[] findBall(int[][] grid) { int columns = grid[0].length; int[] bottoms = new int[columns]; for (int i = 0; i < columns; i++) bottoms[i] = findBottom(grid, i); return bottoms; } public int findBottom(int[][] grid, int start) { int rows = grid.length, columns = grid[0].length; int row = 0, column = start; while (row < rows) { int curr = grid[row][column]; if (curr == 1) column++; else column; if (column < 0  column >= columns) return 1; if (curr != grid[row][column]) return 1; row++; } return column; } }

// OJ: https://leetcode.com/problems/wherewilltheballfall/ // Time: O(MN) // Space: O(N) class Solution { public: vector<int> findBall(vector<vector<int>>& G) { int M = G.size(), N = G[0].size(); vector<int> b(N); iota(begin(b), end(b), 0); for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (b[j] == 1) continue; if (G[i][b[j]] == 1) { if (b[j] == N 1  G[i][b[j] + 1] == 1) b[j] = 1; else b[j]++; } else { if (b[j] == 0  G[i][b[j]  1] == 1) b[j] = 1; else b[j]; } } } return b; } };

# 1706. Where Will the Ball Fall # https://leetcode.com/problems/wherewilltheballfall/ class Solution: def findBall(self, grid: List[List[int]]) > List[int]: res = [] for c in range(len(grid[0])): res.append(self.helper(grid, 0, c)) return res def helper(self, grid, r, c): if r == len(grid): return c if r >= 0 and r < len(grid) and c >= 0 and c < len(grid[r]): if c == 0 and grid[r][c] == 1: return 1 elif c == len(grid[r])  1 and grid[r][c] == 1: return 1 elif c < len(grid[r])  1 and grid[r][c] == 1 and grid[r][c+1] == 1: return 1 elif c > 0 and grid[r][c1] == 1 and grid[r][c] == 1: return 1 return self.helper(grid, r+1, c + grid[r][c]) return 1