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:

C++
#include <iostream>
#define M 3
#define N 3
using namespace std;
void rotateMatrix( int matrix[][M], int k) {
int temp[M];
k = k % M;
for ( int i = 0; i < N; i++) {
for ( int t = 0; t < M - k; t++)
temp[t] = matrix[i][t];
for ( int j = M - k; j < M; j++)
matrix[i][j - M + k] = matrix[i][j];
for ( int j = k; j < M; j++)
matrix[i][j] = temp[j - k];
}
}
void displayMatrix( int matrix[][M]) {
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < M; j++)
cout << matrix[i][j] << " " ;
cout << endl;
}
}
int main() {
int matrix[N][M] = {{12, 23, 34},
{45, 56, 67},
{78, 89, 91}};
int k = 2;
rotateMatrix(matrix, k);
displayMatrix(matrix);
return 0;
}
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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!