Coding Learning Daily Challenge - 1/11/2022 - 1706. Where Will the Ball Fall

Q: 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 i^th column at the top, or -1 if the ball gets stuck in the box.

* [grid[0].length] = column number

1. Depth First Search (DFS):

Condition 1: The ball reaches a point where it can no longer move ahead. In this case, we will return -1. Condition 2: The ball reaches the last row. In this case, we will return the column in which the ball will drop.

*Using Recursive function

class Solution {

public int[] findBall(int[][] grid) {

// create new int[], for int[grid[0].length]

int result[] = new int[grid[0].length];

// how many ball as well as how many row

for (int i = 0; i < grid[0].length; i++) {

result[i] = findBallDropColumn(0, i, grid);

}

return result;

}

public int findBallDropColumn(int row, int col, int[][] grid) {

// base case; ball reached the last row

if (row == grid.length)

return col;

int nextColumn = col + grid[row][col];

if (nextColumn < 0 ||

nextColumn > grid[0].length - 1 ||

// [1,1] & [-1,-1]drop, [1,-1] & [-1,1] stuck

grid[row][col] != grid[row][nextColumn]) {

return -1;

}

return findBallDropColumn(row + 1, nextColumn, grid);

}

2. Dynamic Programming Approach:

class Solution {

public int[] findBall(int[][] grid) {

int result[] = new int[grid[0].length];

Integer memo[][] = new Integer[grid.length + 1][grid[0].length];404Please respect copyright.ＰＥＮＡＮＡboOdupmQli

404Please respect copyright.ＰＥＮＡＮＡmI1ZMoazp3

for (int row = grid.length; row >= 0; row--) {

for (int col = 0; col < grid[0].length; col++) {

if (row == grid.length) {

// for the last row 404Please respect copyright.ＰＥＮＡＮＡzUxXKzCsvs

memo[row][col] = col;

continue;

}

// for the remaining row.

int nextColumn = col + grid[row][col];

if (nextColumn < 0 ||

nextColumn > grid[0].length - 1 ||

grid[row][col] != grid[row][nextColumn])

memo[row][col] = -1;

else

memo[row][col] = memo[row + 1][nextColumn];

// reaching row 0, copy the values in all the column in the result array. 404Please respect copyright.ＰＥＮＡＮＡjvNzEon3qL

if (row == 0) {

result[col] = memo[row][col];

}

return result;

}

3. Iterative Approach, Simulation:

class Solution {

public int[] findBall(int[][] grid) {

int result[] = new int[grid[0].length];

// 1. Start by picking up a ball starting from every column col, iterating from the 0th column to the last column. Initialize the current column as col.

for (int col = 0; col < grid[0].length; col++) {

int currentCol = col;

// 2. For every such ball, simulate its movement in every row ranging from the 0th to the nth row.

for (int row = 0; row < grid.length; row++) {

int nextColumn = currentCol + grid[row][currentCol];

// stuck case

if (nextColumn < 0 ||

nextColumn > grid[0].length - 1 ||

grid[row][currentCol] != grid[row][nextColumn]) {

result[col] = -1;

break;

}

// 3. Update the value of nextColumn in the result array for every row. In the end, the result will store the column number where the ball will fall after the last row. (result[col] = currentCol, return the result)

result[col] = nextColumn;

currentCol = nextColumn;

}

return result;

}

ns216.73.216.30da2

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