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Rotation of Matrix

ISC 2015 Computer Science Practical, Question 2 · Arrays and matrices

The question

Write a program to declare a square matrix A[][] of order M x M where ‘M’ is the number of rows and the number of columns, such that M must be greater than 2 and less than 10. Accept the value of M as user input. Display an appropriate message for an invalid input. Allow the user to input integers into the matrix. Perform the following tasks: a) Display the original matrix. b) Rotate the matrix 90o clockwise as shown below: Original matrix Rotated matrix 1 2 3 7 4 1 4 5 6 8 5 2 7 8 9 9 6 3

c) Find the sum of the elements of the four corners of the matrix.

Test your program for the following data and some random data:
Example 1:
INPUT:  M = 3
        3   4   9
        2   5   8
        1   6   7
OUTPUT:
ORIGINAL MATRIX
        3   4   9
        2   5   8
        1   6   7
MATRIX AFTER ROTATION
        1   2   3
        6   5   4
        7   8   9
Sum of the corner elements = 20

Example 2:
INPUT:  M = 4
        1   2   4   9
        2   5   8   3
        1   6   7   4
        3   7   6   5
OUTPUT:
ORIGINAL MATRIX
        1   2   4   9
        2   5   8   3
        1   6   7   4
        3   7   6   5
MATRIX AFTER ROTATION
        3   1   2   1
        7   6   5   2
        6   7   8   4
        5   4   3   9
Sum of the corner elements = 18

EXAMPLE 3
INPUT:  M = 14
OUTPUT: SIZE OUT OF RANGE

Question: ISC 2015 Computer Science Paper 2 (Practical), Question 2, set by the Council for the Indian School Certificate Examinations (CISCE). Solution and video: Vinay Singh.

Java solution

ISC2015Q2.java Download GitHub

import java.util.*;

class Matrix {
    private int A[][], M;

    public Matrix(int M) {
        A = new int[M][M];
        this.M = M;
    }

    public void getInput() {
        Scanner in = new Scanner(System.in);
        for (int row = 0; row < M; row++) {
            for (int col = 0; col < M; col++) {
                A[row][col] = in.nextInt();
            }
        }
        in.close();
    }

    public String toString() {
        String output = "";
        for (int row = 0; row < M; row++) {
            output += "\t";
            for (int col = 0; col < M; col++) {
                output += A[row][col];
                if (col < M)
                    output += "\t";
            }
            output += "\n";
        }
        return output;
    }

    public void rotate() {
        int half = M / 2, temp;
        for (int i = 0; i < half; i++) {
            for (int x = i; x < M - i - 1; x++) {
                temp = A[M - x - 1][i];
                A[M - x - 1][i] = A[M - i - 1][M - x - 1];
                A[M - i - 1][M - x - 1] = A[x][M - i - 1];
                A[x][M - i - 1] = A[i][x];
                A[i][x] = temp;
            }
        }
    }
    public int getCornerSum(){
        return A[0][0]+A[0][M-1]+A[M-1][M-1]+A[M-1][0];
    }
}

public class ISC2015Q2 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        System.out.print("INPUT:\tM = ");
        int M = in.nextInt();
        if (M > 2 && M < 10) {
            Matrix objMatrix = new Matrix(M);
            objMatrix.getInput();
            System.out.println("OUTPUT:");
            System.out.println("ORIGINAL MATRIX");
            System.out.println(objMatrix);
            objMatrix.rotate();
            System.out.println("MATRIX AFTER ROTATION");
            System.out.println(objMatrix);
            System.out.println("Sum of the corner elements = "+objMatrix.getCornerSum());
        } else {
            System.out.println("OUTPUT: SIZE OUT OF RANGE");
        }
        in.close();
    }
}

Rotation of Matrix

The question

Java solution

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