Java Program to Inplace rotate square matrix by 90 degrees | Set 1
Given a square matrix, turn it by 90 degrees in anti-clockwise direction without using any extra space.
Examples :
Input:
Matrix:
1 2 3
4 5 6
7 8 9
Output:
3 6 9
2 5 8
1 4 7
The given matrix is rotated by 90 degree
in anti-clockwise direction.
Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
The given matrix is rotated by 90 degree
in anti-clockwise direction.
An approach that requires extra space is already discussed here.
Approach: To solve the question without any extra space, rotate the array in form of squares, dividing the matrix into squares or cycles. For example,
A 4 X 4 matrix will have 2 cycles. The first cycle is formed by its 1st row, last column, last row and 1st column. The second cycle is formed by 2nd row, second-last column, second-last row and 2nd column. The idea is for each square cycle, swap the elements involved with the corresponding cell in the matrix in anti-clockwise direction i.e. from top to left, left to bottom, bottom to right and from right to top one at a time using nothing but a temporary variable to achieve this.
Demonstration:
First Cycle (Involves Red Elements)
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Moving first group of four elements (First
elements of 1st row, last row, 1st column
and last column) of first cycle in counter
clockwise.
4 2 3 16
5 6 7 8
9 10 11 12
1 14 15 13
Moving next group of four elements of
first cycle in counter clockwise
4 8 3 16
5 6 7 15
2 10 11 12
1 14 9 13
Moving final group of four elements of
first cycle in counter clockwise
4 8 12 16
3 6 7 15
2 10 11 14
1 5 9 13
Second Cycle (Involves Blue Elements)
4 8 12 16
3 6 7 15
2 10 11 14
1 5 9 13
Fixing second cycle
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
Algorithm:
- There is N/2 squares or cycles in a matrix of side N. Process a square one at a time. Run a loop to traverse the matrix a cycle at a time, i.e loop from 0 to N/2 – 1, loop counter is i
- Consider elements in group of 4 in current square, rotate the 4 elements at a time. So the number of such groups in a cycle is N – 2*i.
- So run a loop in each cycle from x to N – x – 1, loop counter is y
- The elements in the current group is (x, y), (y, N-1-x), (N-1-x, N-1-y), (N-1-y, x), now rotate the these 4 elements, i.e (x, y) <- (y, N-1-x), (y, N-1-x)<- (N-1-x, N-1-y), (N-1-x, N-1-y)<- (N-1-y, x), (N-1-y, x)<- (x, y)
- Print the matrix.
Java
import java.io.*;
class GFG {
static void rotateMatrix(
int N, int mat[][])
{
for ( int x = 0 ; x < N / 2 ; x++) {
for ( int y = x; y < N - x - 1 ; y++) {
int temp = mat[x][y];
mat[x][y] = mat[y][N - 1 - x];
mat[y][N - 1 - x]
= mat[N - 1 - x][N - 1 - y];
mat[N - 1 - x][N - 1 - y] = mat[N - 1 - y][x];
mat[N - 1 - y][x] = temp;
}
}
}
static void displayMatrix(
int N, int mat[][])
{
for ( int i = 0 ; i < N; i++) {
for ( int j = 0 ; j < N; j++)
System.out.print(
" " + mat[i][j]);
System.out.print("
");
}
System.out.print("
");
}
public static void main(String[] args)
{
int N = 4 ;
int mat[][] = {
{ 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 9 , 10 , 11 , 12 },
{ 13 , 14 , 15 , 16 }
};
rotateMatrix(N, mat);
displayMatrix(N, mat);
}
}
|
Output :
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
Complexity Analysis:
- Time Complexity: O(n*n), where n is size of array.
A single traversal of the matrix is needed.
- Space Complexity: O(1).
As a constant space is needed
Please refer complete article on Inplace rotate square matrix by 90 degrees | Set 1 for more details!
Last Updated :
17 Aug, 2023
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