Given a square matrix mat[][] of dimension N and an integer K, the task is to rotate the matrix by 90 degrees K times without changing the position of the diagonal elements.
Examples:
Input: mat[][] = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}}, K = 1
Output:
 1 16 11 6 5 Â
 22 7 12 9 2
 23 18 13 8 3
 24 17 14 19 4
 21 20 15 10 25Input: mat[][] = {{10, 11}, {12, 13}}, K = 2
Output:
10 11
12 13
Approach: The given problem can be solved by using the idea discussed in this article and the fact that the matrix restores after performing clockwise rotation 4 times. Follow the below steps to solve the given problem:
- Update the value of K as K % 4.
- Iterate until K is a positive and perform the following steps:
- Traverse the matrix, for i over the range [0, N / 2) and j over the range[0, N – i – 1) and perform the following steps:
- If the value of i != j and (i + j) != (N – 1), then perform the following steps:
- Store the value of mat[i][j] in a temporary variable temp.
- Update the value of mat[i][j] as mat[N – 1 – j][i].
- Update the value of mat[N – 1 – j][i] as mat[N – 1 -i][N – 1 – j].
- Update the value of mat[N – 1  – i][N – 1 – j] as mat[j][N – 1 – i].
- Update the value of mat[j][N – 1 – i] as temp.
- After completing the above steps, print the updated matrix obtained.
Below is the implementation of the above approach:
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;Â
public class GFG {Â
    // Function to print the matrix    static void print(int mat[][])    {        // Iterate over the rows        for (int i = 0; i < mat.length; i++) {Â
            // Iterate over the columns            for (int j = 0; j < mat[0].length; j++)Â
                // Print the value                System.out.print(mat[i][j] + " ");                       System.out.println();        }    }Â
    // Function to perform the swapping of    // matrix elements in clockwise manner    static void performSwap(int mat[][], int i, int j)    {        int N = mat.length;Â
        // Stores the last row        int ei = N - 1 - i;Â
        // Stores the last column        int ej = N - 1 - j;Â
        // Perform the swaps        int temp = mat[i][j];        mat[i][j] = mat[ej][i];        mat[ej][i] = mat[ei][ej];        mat[ei][ej] = mat[j][ei];        mat[j][ei] = temp;    }Â
    // Function to rotate non - diagonal    // elements of the matrix K times in    // clockwise direction    static void rotate(int mat[][], int N, int K)    {        // Update K to K % 4        K = K % 4;Â
        // Iterate until K is positive        while (K-- > 0) {Â
            // Iterate each up to N/2-th row            for (int i = 0; i < N / 2; i++) {Â
                // Iterate each column                // from i to N - i - 1                for (int j = i; j < N - i - 1; j++) {Â
                    // Check if the element                    // at i, j is not a                    // diagonal element                    if (i != j && (i + j) != N - 1) {Â
                        // Perform the swapping                        performSwap(mat, i, j);                    }                }            }        }Â
        // Print the matrix        print(mat);    }       // Driver Code    public static void main(String[] args)    {Â
        int K = 5;        int mat[][] = {            { 1, 2, 3, 4 },            { 6, 7, 8, 9 },            { 11, 12, 13, 14 },            { 16, 17, 18, 19 },        };               int N = mat.length;        rotate(mat, N, K);    }}Â
// This code is contributed by Kingash. |
1 11 6 4 17 7 8 2 18 12 13 3 16 14 9 19
Â
Time Complexity: O(N3) since using three inner loop one while and two for loop
Auxiliary Space: O(1)
Please refer complete article on Rotate all Matrix elements except the diagonal K times by 90 degrees in clockwise direction for more details!
