Given a matrix of size N*M, and a number K. We have to rotate the matrix K times to the right side.
Examples:
Input : N = 3, M = 3, K = 2 12 23 34 45 56 67 78 89 91 Output : 23 34 12 56 67 45 89 91 78 Input : N = 2, M = 2, K = 2 1 2 3 4 Output : 1 2 3 4
A simple yet effective approach is to consider each row of the matrix as an array and perform an array rotation. This can be done by copying the elements from K to end of array to starting of array using temporary array. And then the remaining elements from start to K-1 to end of the array.
Lets take an example:
Java
// Java program to rotate a matrix // right by k times class GFG
{ // size of matrix
static final int M= 3 ;
static final int N= 3 ;
// function to rotate matrix by k times
static void rotateMatrix( int matrix[][], int k)
{
// temporary array of size M
int temp[]= new int [M];
// within the size of matrix
k = k % M;
for ( int i = 0 ; i < N; i++)
{
// copy first M-k elements
// to temporary array
for ( int t = 0 ; t < M - k; t++)
temp[t] = matrix[i][t];
// copy the elements from k
// to end to starting
for ( int j = M - k; j < M; j++)
matrix[i][j - M + k] = matrix[i][j];
// copy elements from
// temporary array to end
for ( int j = k; j < M; j++)
matrix[i][j] = temp[j - k];
}
}
// function to display the matrix
static void displayMatrix( int matrix[][])
{
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < M; j++)
System.out.print(matrix[i][j] + " " );
System.out.println();
}
}
// Driver code
public static void main (String[] args)
{
int matrix[][] = {{ 12 , 23 , 34 },
{ 45 , 56 , 67 },
{ 78 , 89 , 91 }};
int k = 2 ;
// rotate matrix by k
rotateMatrix(matrix, k);
// display rotated matrix
displayMatrix(matrix);
}
} // This code is contributed by Anant Agarwal. |
Output:
23 34 12 56 67 45 89 91 78
Time Complexity: O(N*M)
Auxiliary Space: O(M)
Please refer complete article on Rotate the matrix right by K times for more details!