Question

In Java please You are given a matrix of characters representing a big box. Each cell of the matrix contains one of three characters: " / "which means that the cell is empty; '*', which means that the cell contains an obstacle; '+', which means that the cell contains a small box. You decide to rotate the big box clockwise to see how the small boxes will fall under the gravity. After rotating, each small box falls down until it lands on an obstacle, another small box, or the bottom of the big box. Given box, a matrix representation of the big box, your task is to return the state of the box after rotating it clockwise.

Answer 1

CODE:

package bigbox;

import java.util.Scanner;

public class BigBox
{
    public static void main(String[] args)
    {
        //variable to store number of rows and columns and taking it input from user
        int row,col;
        Scanner scan = new Scanner(System.in);
        System.out.print("Enter the number of rows: ");
        row = scan.nextInt();
        System.out.print("Enter the number of columns: ");
        col = scan.nextInt();
        //defining a char array of the given size
        char[][] arr = new char[row][col];
        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++)
                arr[i][j]=scan.next().charAt(0);
        //calling the rotate big box function to rotate the box
        rotateBigBox(arr,row,col);
        //printing the new big box
        for(int i=0;i<row;i++)
        {
            System.out.println();
            for(int j=0;j<col;j++)
                System.out.print(arr[i][j]);
        }
    }
    //this function rotates the big box
    private static void rotateBigBox(char[][] arr, int row, int col)
    {
        //for loop to iterate rows
        for(int i=0;i<row;i++)
        {   //for loop to iterate column by colum value of a row from right to left
            for(int j=col-2;j>=0;j--)
            {
                //if we find a small box
                if(arr[i][j]=='+')
                {
                    // we define a variable 'k' which is 1 more than curent 'j'
                    int k=j+1;
                    //we iterate using k till we reach end of the big box and till we are finding '/' at the k-th place
                    while(k<col && arr[i][k]=='/')
                    {
                        //we swap if we find '/' in k-th place.
                        if(arr[i][k]=='/')
                        {
                            arr[i][k]='+';
                            arr[i][k-1]='/';
                        }
                        //we increment value of k
                        k++;
                    }
                }
            }
        }
        //rotating the box clockwise
        for (int i = 0; i < row / 2; i++)
        {
            for (int j = i; j < col - i - 1; j++)
            {
                char temp = arr[i][j];
                arr[i][j] = arr[row - 1 - j][i];
                arr[row - 1 - j][i] = arr[row - 1 - i][col - 1 - j];
                arr[row - 1 - i][row - 1 - j] = arr[j][col - 1 - i];
                arr[j][row - 1 - i] = temp;
            }
        }
    }  
}

OUTPUT:

Enter the number of rows: 4
Enter the number of columns: 4
+ / + /
+ / * +
* * + /
+ + + /

/ * / /
+ * + /
+ / * +
+ + + +

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