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1706. Where Will the Ball Fall

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.

 

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.

Solutions

Solution 1: Case Analysis + DFS

We can use DFS to simulate the movement of the ball. Design a function \(\textit{dfs}(i, j)\), which represents the column where the ball will fall when it starts from row \(i\) and column \(j\). The ball will get stuck in the following cases:

  1. The ball is in the leftmost column, and the cell's diagonal directs the ball to the left.
  2. The ball is in the rightmost column, and the cell's diagonal directs the ball to the right.
  3. The cell's diagonal directs the ball to the right, and the adjacent cell to the right directs the ball to the left.
  4. The cell's diagonal directs the ball to the left, and the adjacent cell to the left directs the ball to the right.

If any of the above conditions are met, we can determine that the ball will get stuck and return \(-1\). Otherwise, we can continue to recursively find the next position of the ball. Finally, if the ball reaches the last row, we can return the current column index.

The time complexity is \(O(m \times n)\), and the space complexity is \(O(m)\). Here, \(m\) and \(n\) are the number of rows and columns of the grid, respectively.

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class Solution:
    def findBall(self, grid: List[List[int]]) -> List[int]:
        def dfs(i: int, j: int) -> int:
            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)

        m, n = len(grid), len(grid[0])
        return [dfs(0, j) for j in range(n)]

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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);
    }
}

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class Solution {
public:
    vector<int> findBall(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<int> ans(n);
        function<int(int, 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);
        };
        for (int j = 0; j < n; ++j) {
            ans[j] = dfs(0, j);
        }
        return ans;
    }
};

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func findBall(grid [][]int) (ans []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)
    }
    for j := 0; j < n; j++ {
        ans = append(ans, dfs(0, j))
    }
    return
}

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function findBall(grid: number[][]): number[] {
    const m = grid.length;
    const n = grid[0].length;
    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);
        }
    };
    return Array.from({ length: n }, (_, j) => dfs(0, j));
}

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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 ans = vec![0; n];
        for i in 0..n {
            ans[i] = Self::dfs(&grid, 0, i);
        }
        ans
    }
}

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/**
 * @param {number[][]} grid
 * @return {number[]}
 */
var findBall = function (grid) {
    const m = grid.length;
    const n = grid[0].length;
    const dfs = (i, j) => {
        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);
        }
    };
    return Array.from({ length: n }, (_, j) => dfs(0, j));
};

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