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:
Javascript
<script>// Javascript program to rotate a matrix // right by k times // size of matrix var M = 3; var N = 3; // function to rotate matrix by k times function rotateMatrix(matrix , k) { // temporary array of size M var temp = Array(M).fill(0); // within the size of matrix k = k % M; for (i = 0; i < N; i++) { // copy first M-k elements // to temporary array for (t = 0; t < M - k; t++) temp[t] = matrix[i][t]; // copy the elements from k // to end to starting for (j = M - k; j < M; j++) matrix[i][j - M + k] = matrix[i][j]; // copy elements from // temporary array to end for (j = k; j < M; j++) matrix[i][j] = temp[j - k]; } } // function to display the matrix function displayMatrix(matrix) { for (i = 0; i < N; i++) { for (j = 0; j < M; j++) document.write(matrix[i][j] + " "); document.write("<br/>"); } } // Driver code var matrix = [ [ 12, 23, 34 ], [ 45, 56, 67 ], [ 78, 89, 91 ] ]; var k = 2; // rotate matrix by k rotateMatrix(matrix, k); // display rotated matrix displayMatrix(matrix);// This code contributed by umadevi9616 </script> |
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!
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