Where Will the Ball Fall

Problem Description

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.

Examples

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

Solution for 1706. Where Will the Ball Fall

Approach

We drop the ball at grid[0][i] and track ball position i1, which initlized as i.

An observation is that, if the ball wants to move from i1 to i2, we must have the board direction grid[j][i1] == grid[j][i2]

Code in Different Languages

Python

Written by @agarwalhimanshugaya

        def findBall(self, grid):
     m, n = len(grid), len(grid[0])

     def test(i):
         for j in xrange(m):
             i2 = i + grid[j][i]
             if i2 < 0 or i2 >= n or grid[j][i2] != grid[j][i]:
                 return -1
             i = i2
         return i
     return map(test, range(n))

Java

Written by @agarwalhimanshugaya

       public int[] findBall(int[][] grid) {
     int m = grid.length, n = grid[0].length, res[] = new int[n];
     for (int i = 0; i < n; ++i) {
         int i1 = i, i2;
         for (int j = 0; j < m; ++j) {
             i2 = i1 + grid[j][i1];
             if (i2 < 0 || i2 >= n || grid[j][i2] != grid[j][i1]) {
                 i1 = -1;
                 break;
             }
             i1 = i2;
         }
         res[i] = i1;
     }
     return res;
 }

C++

Written by @agarwalhimanshugaya

       vector<int> findBall(vector<vector<int>>& grid) {
     int m = grid.size(), n = grid[0].size();
     vector<int> res;
     for (int i = 0; i < n; ++i) {
         int i1 = i, i2;
         for (int j = 0; j < m; ++j) {
             i2 = i1 + grid[j][i1];
             if (i2 < 0 || i2 >= n || grid[j][i2] != grid[j][i1]) {
                 i1 = -1;
                 break;
             }
             i1 = i2;
         }
         res.push_back(i1);
     }
     return res;
 }

Complexity Analysis

• Time Complexity: $O(N*M)$
• Space Complexity: $O(N)$

References

• LeetCode Problem: Where Will the Ball Fall
• Solution Link: LeetCode Solution