Where Will the Ball Fall

class Solution {
public:
    vector<int> trackBalls(vector<vector<int>>& matrix) {
        vector<int> response(matrix[0].size(), 0);
    
        for (int columnIndex = 0; columnIndex < matrix[0].size(); columnIndex++) {
            int pos = columnIndex;
            for (int rowIndex = 0; rowIndex < matrix.size(); rowIndex++) {
                int newPos = pos + matrix[rowIndex][pos];
                if (newPos < 0 || newPos >= matrix[0].size() ||
                    matrix[rowIndex][pos] != matrix[rowIndex][newPos]) {
                    response[columnIndex] = -1;
                    break;
                }
                response[columnIndex] = newPos;
                pos = newPos;
            }
        }
        return response;
    }
};

The given problem is about predicting the exit column for balls that are dropped into a grid, where each cell of the grid either causes the ball to move right or left. The solution is implemented in C++ within a class named Solution.

• The function trackBalls takes a 2D vector matrix as an input. This matrix represents the grid where balls are dropped.

• For each initial column index, simulate the path of the ball through the rows of the grid. Depending on the value at the current position in the grid, move the ball to a new position ('1' indicates a move to the right and '-1' a move to the left).

• Ensure checks are in place:

  • The ball should not move outside the grid boundaries.
  • The ball can only move to a new position if it continues in the same direction indicated by the current and new position values.

• If a condition is violated (i.e., the ball moves out of grid boundaries, or attempts to switch direction incorrectly), set the exit column for that ball to -1 indicating that the ball will be stuck in the grid or fall out.

• If all conditions are met successfully for a column, record the final position as the exit column for the ball.

• Return the result in a vector response that documents the exit column for each ball corresponding to its start column.

This solution ensures that the ball’s path is tracked correctly across the grid based on the directional constraints, and it efficiently handles boundary and direction conditions that determine the successful exit or failure points for each ball.

class Solution {
    public int[] simulateBallDrop(int[][] maze) {
        int[] outcomes = new int[maze[0].length];
    
        for (int startPosition = 0; startPosition < maze[0].length; startPosition++) {
            int currentPos = startPosition;
            for (int level = 0; level < maze.length; level++) {
                int newPos = currentPos + maze[level][currentPos];
                if (newPos < 0 || newPos >= maze[0].length || maze[level][currentPos] != maze[level][newPos]) {
                    outcomes[startPosition] = -1;
                    break;
                }
                outcomes[startPosition] = newPos;
                currentPos = newPos;
            }
        }
        return outcomes;
    }
}

The Java program provided defines a method named simulateBallDrop in a class named Solution. This method is designed to simulate the fall of balls through a vertical maze specified by an array of integers, where each element in the array can either be 1 (indicating the ball can roll to the right) or -1 (indicating the ball can roll to the left).

• The maze is represented by a two-dimensional integer array named maze.
• The method returns an array of integers representing the final positions of the balls. A ball's position is determined from its starting position in the top row of the maze.
• For each starting position on the top row:
  1. Determine the new position of the ball at each level of the maze.
  2. If the new position causes the ball to exit the boundaries of the maze (either on the left or right) or results in a mismatch between the entry and exit directions at any level (meaning the next move in the direction pushed by the maze is blocked or doesn't exist), mark the outcome for that starting position as -1 to indicate the ball has fallen out.
  3. Continue simulating the ball's drop until it either falls out or reaches the last level.

The for loop iterates through each possible starting position (column) of the maze, and for each starting position, it iterates downward through each row (level) of the maze. The current position of the ball updates depending on the cell's value in the maze (maze[level][currentPos]). If the ball remains within the grid and the directions remain consistent, the ball's final position for that column is recorded.

Ultimately, the method returns an array (outcomes) where each element corresponds to the final position of a ball that starts in the respective column of the first row or -1 if the ball does not exit the maze properly.