Rotate Image - Java - algorithm

Question

I am doing a question where, given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation. This my my code:
class Solution {
public void rotate(int[][] matrix) {
int size = matrix.length;
for(int i = 0 ; i < matrix.length; i++){
for(int y = 0 ; y < matrix[0].length ; y++){
matrix[i][y] = matrix[size - y - 1][i];
System.out.println(size - y - 1);
System.out.println(i);
System.out.println("");
}
}
}
}
This is the input and output results:
input matrix: [[1,2,3],[4,5,6],[7,8,9]]
output matrix: [[7,4,7],[8,5,4],[9,4,7]]
expected matrix: [[7,4,1],[8,5,2],[9,6,3]]
I do not really understand why I am getting duplicates in my output such as the number seven 3 times. On my System.out.println statement, I am getting the correct list of indexes :
2
0
1
0
0
0
2
1
1
1
0
1
2
2
What can be wrong?

Answer 1

I have found a solution. I will try my best to explain it.
Let us consider an array of size 4.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Now lets look at the numbers present only on the outside of the array:
1 2 3 4
5 8
9 12
13 14 15 16
We will proceed by storing the first element 1 in a temporary variable. Next we will replace 1 by 13, 13 by 16, 16 by 4 and at last 4 by 1 (whose value we already stored in the temporary variable).
We will do the same for all the elements of the first row.
Here is a pseudocode if you just want to rotate this outer ring, lets call it an outer ring:
for i = 0 to n-1
{
temp = A[0][i];
A[0][i] = A[n-1-i][0];
A[n-1-i][0] = A[n-1-0][n-1-i];
A[n-1-0][n-1-i] = A[i][n-1-0];
A[i][n-1-0] = temp;
}
The code runs for a total of n times. Once for each element of first row. Implement this code an run it. You will see only the outer ring is rotated. Now lets look at the inner ring:
6 7
10 11
Now the loop in pseudocode only needs to run for 2 times and also our range of indexes has decreased. For outer ring, the loop started from i = 0 and ended at i = n-1. However, for the inner ring the for loop need to run from i = 1 to i = n-2.
If you had an array of size n, to rotate the xth ring of the array, the loop needs to run from i = x to i = n-1-x.
Here is the code to rotate the entire array:
x = 0;
int temp;
while (x < n/2)
{
for (int i = x;i < n-1-x;i++)
{
temp = arr[x][i];
arr[x][i] = arr[n-1-i][x];
arr[n-1-i][x] = arr[n-1-x][n-1-i];
arr[n-1-x][n-1-i] = arr[i][n-1-x];
arr[i][n-1-x] = temp;
}
x++;
}
Here each value of x denotes the xth ring.
0 <= x <= n-1
The reason why the outer loop runs only for x < n/2 times is because each array has n/2 rings when n is even and n/2 + 1 rings if n is odd.
I hope I have helped you. Do comment if face any problems with the solution or its explanation.