## Problem Statement :

You are 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.

## Test Cases :

Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]

Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]

Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]

Example 3:

Input: matrix = [[1]]

Output: [[1]]

Example 4:

Input: matrix = [[1,2],[3,4]]

Output: [[3,1],[4,2]]

## Constraints:

matrix.length == n

matrix[i].length == n

1 <= n <= 20

-1000 <= matrix[i][j] <= 1000

## Algorithm :

The simple idea is, consider this matrix

```
matrix = [[1,2,3],[4,5,6],[7,8,9]]
```

```
1 2 3
4 5 6
7 8 9
Just take the transpose and lets see what we got :
1 2 3 1 4 7
4 5 6. => 2 5 8
7 8 9 3 6 9
So we got the above after doing the transpose operation.
```

```
for (int i=0; i<row; i++) {
for (int j=0; j<i; j++) {
swap(matrix, i, j);
}
}
```

```
Now let's compare it with our possible answer.
Our transpose is :
1 4 7 7 4 1
2 5 8. and our answer is 8 5 2
3 6 9 9 6 3
So what we observe, answer is just reversing each row after doing the transpose.
1 4 7. after reversing 7 4 1
2 5 8 after reversing 8 5 2
3 6 9 after reversing 9 6 3
which is what we needed as our answer.
```

So algorithm is pretty simple :

1) Take the transpose of the given matrix

2) Reverse each row

```
Note : The above is 90 degree clockwise direction. What if, we need to rotate 90 degree anti clockwise direction.
Same way, just take the transpose.
1 2 3 1 4 7
4 5 6. => 2 5 8
7 8 9 3 6 9
Now what will be our possible answer if we do anti clockwise 90 degree rotation on given matrix.
1 2 3 3 6 9
4 5 6. => 2 5 8
7 8 9 1 4 7
Now we can just observe the answer along with the transpose of the matrix
1 4 7 3 6 9
2 5 8. and our answer is 2 5 8
3 6 9 1 4 7
So we can observe that if we just reverse all the column, we will get the desired output.
1 4 7. after reversing column wise 3 6 9
2 5 8 2 5 8
3 6 9 1 4 7
```

## Complexity :

Time complexity is O(row * col) and since we are doing in-place, space complexity is O(1).

## Code : (Clockwise 90 degree rotation)

```
class Solution {
public void rotate(int[][] matrix) {
if (matrix == null || matrix.length == 0) {
return;
}
int row = matrix.length;
int col = matrix[0].length;
for (int i=0; i<row; i++) {
for (int j=0; j<i; j++) {
swap(matrix, i, j);
}
}
for (int [] rowWise : matrix) {
reverse(rowWise, 0, col - 1);
}
}
private void swap(int [][] matrix, int i, int j) {
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
private void reverse(int [] nums, int i, int j) {
while (i < j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
i++;
j--;
}
}
}
```

## Github :

## Rohithv07 / LeetCodeTopInterviewQuestions

### Leetcode Top Interview questions discussed in Leetcode. https://leetcode.com/explore/interview/card/top-interview-questions

# LeetCodeTopInterviewQuestions

Leetcode Top Interview questions discussed in Leetcode. https://leetcode.com/explore/interview/card/top-interview-questions-easy/

Also Question answered from CodeSignal.com : https://app.codesignal.com/

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