Given a matrix mat[][] of size N*N, the task is to rotate the matrix by 45 degrees and print the matrix.
Examples:
Input: N = 6,
mat[][] = {{3, 4, 5, 1, 5, 9, 5},
{6, 9, 8, 7, 2, 5, 2},
{1, 5, 9, 7, 5, 3, 2},
{4, 7, 8, 9, 3, 5, 2},
{4, 5, 2, 9, 5, 6, 2},
{4, 5, 7, 2, 9, 8, 3}}
Output:
3
6 4
1 9 5
4 5 8 1
4 7 9 7 5
4 5 8 7 2 9
5 2 9 5 5
7 9 3 3
2 5 5
9 6
8Input: N = 4,
mat[][] = {{2, 5, 7, 2},
{9, 1, 4, 3},
{5, 8, 2, 3},
{6, 4, 6, 3}}Output:
2
9 5
5 1 7
6 8 4 2
4 2 3
6 3
3
Approach: Follow the steps given below in order to solve the problem:
- Store the diagonal elements in a list using a counter variable.
- Print the number of spaces required to make the output look like the desired pattern.
- Print the list elements after reversing the list.
- Traverse through only diagonal elements to optimize the time taken by the operation.
Below is the implementation of the above approach:
Java
// Java program for// the above approachimport java.util.*;class GFG{// Function to rotate// matrix by 45 degreestatic void matrix(int n, int m, int [][]li){ // Counter Variable int ctr = 0; while (ctr < 2 * n - 1) { for(int i = 0; i < Math.abs(n - ctr - 1); i++) { System.out.print(" "); } Vector<Integer> lst = new Vector<Integer>(); // Iterate [0, m] for(int i = 0; i < m; i++) { // Iterate [0, n] for(int j = 0; j < n; j++) { // Diagonal Elements // Condition if (i + j == ctr) { // Appending the // Diagonal Elements lst.add(li[i][j]); } } } // Printing reversed Diagonal // Elements for(int i = lst.size() - 1; i >= 0; i--) { System.out.print(lst.get(i) + " "); } System.out.println(); ctr += 1; }}// Driver code public static void main(String[] args){ // Dimensions of Matrix int n = 8; int m = n; // Given matrix int[][] li = {{4, 5, 6, 9, 8, 7, 1, 4}, {1, 5, 9, 7, 5, 3, 1, 6}, {7, 5, 3, 1, 5, 9, 8, 0}, {6, 5, 4, 7, 8, 9, 3, 7}, {3, 5, 6, 4, 8, 9, 2, 1}, {3, 1, 6, 4, 7, 9, 5, 0}, {8, 0, 7, 2, 3, 1, 0, 8}, {7, 5, 3, 1, 5, 9, 8, 5}}; // Function call matrix(n, m, li);}}// This code is contributed by Princi Singh |
4
1 5
7 5 6
6 5 9 9
3 5 3 7 8
3 5 4 1 5 7
8 1 6 7 5 3 1
7 0 6 4 8 9 1 4
5 7 4 8 9 8 6
3 2 7 9 3 0
1 3 9 2 7
5 1 5 1
9 0 0
8 8
5
Time Complexity: O(N2)
Auxiliary Space: O(1)
Approach: Rotating Matrix by 45 Degrees using a new matrix
The steps of this approach are:
- Create a new matrix with dimensions (2N-1)x(2N-1), where N is the size of the input matrix. This new matrix will store the rotated matrix.
- Iterate through each element of the input matrix and place it in the correct position in the rotated matrix according to the rotation formula.
- Iterate through each element of the rotated matrix and print it in the correct order, with empty spaces for the elements that were not filled by the input matrix.
Java
public class Main { public static void main(String[] args) { int N = 4; int[][] mat = {{2, 5, 7, 2}, {9, 1, 4, 3}, {5, 8, 2, 3}, {6, 4, 6, 3}}; int[][] result = new int[2*N-1][2*N-1]; // Rotate the matrix by 45 degrees for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { result[i+j][N-1-i+j] = mat[i][j]; } } // Print the rotated matrix for (int i = 0; i < 2*N-1; i++) { for (int j = 0; j < 2*N-1; j++) { if (result[i][j] != 0) { System.out.print(result[i][j] + " "); } else { System.out.print(" "); } } System.out.println(); } }} |
2
9 5
5 1 7
6 8 4 2
4 2 3
6 3
3
The time complexity: O(N^2) and
The Auxiliary space is also O(N^2)
Please refer complete article on Rotate matrix by 45 degrees for more details!
