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Java Program to Rotate Matrix Elements

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Given a matrix, clockwise rotate elements in it.

Examples:

Input
1    2    3
4    5    6
7    8    9

Output:
4    1    2
7    5    3
8    9    6

For 4*4 matrix
Input:
1    2    3    4    
5    6    7    8
9    10   11   12
13   14   15   16

Output:
5    1    2    3
9    10   6    4
13   11   7    8
14   15   16   12

The idea is to use loops similar to the program for printing a matrix in spiral form. One by one rotate all rings of elements, starting from the outermost. To rotate a ring, we need to do following. 
    1) Move elements of top row. 
    2) Move elements of last column. 
    3) Move elements of bottom row. 
    4) Move elements of first column. 
Repeat above steps for inner ring while there is an inner ring.

Below is the implementation of above idea. Thanks to Gaurav Ahirwar for suggesting below solution. 

Java




// Java program to rotate a matrix
import java.lang.*;
import java.util.*;
 
class GFG
{
    static int R = 4;
    static int C = 4;
 
    // A function to rotate a matrix
    // mat[][] of size R x C.
    // Initially, m = R and n = C
    static void rotatematrix(int m,
                    int n, int mat[][])
    {
        int row = 0, col = 0;
        int prev, curr;
 
        /*
        row - Starting row index
        m - ending row index
        col - starting column index
        n - ending column index
        i - iterator
        */
        while (row < m && col < n)
        {
     
            if (row + 1 == m || col + 1 == n)
                break;
     
            // Store the first element of next
            // row, this element will replace
            // first element of current row
            prev = mat[row + 1][col];
     
            // Move elements of first row
            // from the remaining rows
            for (int i = col; i < n; i++)
            {
                curr = mat[row][i];
                mat[row][i] = prev;
                prev = curr;
            }
            row++;
     
            // Move elements of last column
            // from the remaining columns
            for (int i = row; i < m; i++)
            {
                curr = mat[i][n-1];
                mat[i][n-1] = prev;
                prev = curr;
            }
            n--;
     
            // Move elements of last row
            // from the remaining rows
            if (row < m)
            {
                for (int i = n-1; i >= col; i--)
                {
                    curr = mat[m-1][i];
                    mat[m-1][i] = prev;
                    prev = curr;
                }
            }
            m--;
     
            // Move elements of first column
            // from the remaining rows
            if (col < n)
            {
                for (int i = m-1; i >= row; i--)
                {
                    curr = mat[i][col];
                    mat[i][col] = prev;
                    prev = curr;
                }
            }
            col++;
        }
 
            // Print rotated matrix
            for (int i = 0; i < R; i++)
            {
                for (int j = 0; j < C; j++)
                System.out.print( mat[i][j] + " ");
                System.out.print("
");
            }
    }
 
/* Driver program to test above functions */
    public static void main(String[] args)
    {
    // Test Case 1
    int a[][] = { {1, 2, 3, 4},
                  {5, 6, 7, 8},
                {9, 10, 11, 12},
                {13, 14, 15, 16} };
 
    // Tese Case 2
    /* int a[][] = new int {{1, 2, 3},
                            {4, 5, 6},
                            {7, 8, 9}
                        };*/
    rotatematrix(R, C, a);
     
    }
}
 
// This code is contributed by Sahil_Bansall


Output: 

5 1 2 3
9 10 6 4
13 11 7 8
14 15 16 12

Time Complexity: O(max(m,n) * max(m,n))
Auxiliary Space: O(m*n)

Please refer complete article on Rotate Matrix Elements for more details!



Last Updated : 21 Nov, 2022
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