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++
// CPP program to rotate a matrix right by k times #include <iostream> // size of matrix #define M 3 #define N 3 using namespace std;
// function to rotate matrix by k times void rotateMatrix( int matrix[][M], int k) {
// temporary array of size M
int temp[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 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;
}
} // Driver's code int main() {
int matrix[N][M] = {{12, 23, 34},
{45, 56, 67},
{78, 89, 91}};
int k = 2;
// rotate matrix by k
rotateMatrix(matrix, k);
// display rotated matrix
displayMatrix(matrix);
return 0;
} |
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!