1706

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.

 

Constraints:

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

Related Topics:
Dynamic Programming

Solution 1.

// OJ: https://leetcode.com/problems/where-will-the-ball-fall/
// Author: github.com/lzl124631x
// 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;
    }
};