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Medium

Rotate Image

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.

Examples

Input:matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output:[[7,4,1],[8,5,2],[9,6,3]]
The matrix is rotated 90 degrees clockwise.
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]]
The matrix is rotated 90 degrees clockwise.

Constraints

  • n == matrix.length == matrix[i].length
  • 1 <= n <= 20
  • -1000 <= matrix[i][j] <= 1000

Transpose and Reverse (Math Approach)

Approach

Rotating a matrix 90 degrees clockwise can be broken down into two simpler mathematical operations: 1. **Transpose the matrix**: Swap `matrix[i][j]` with `matrix[j][i]`. 2. **Reverse each row**: Swap `matrix[i][j]` with `matrix[i][n - 1 - j]`. This approach is easy to implement and modifies the matrix in-place.

Complexity Analysis

Time Complexity
O(n^2)
Space Complexity
O(1)

Time complexity is O(n^2) because we visit each element in the matrix of size n x n. Space complexity is O(1) since we do it in-place using only temporary variables for swapping.

Solution.java
class Solution {    public void rotate(int[][] matrix) {        int n = matrix.length;                // Step 1: Transpose the matrix        for (int i = 0; i < n; i++) {            for (int j = i; j < n; j++) {                int temp = matrix[i][j];                matrix[i][j] = matrix[j][i];                matrix[j][i] = temp;            }        }                // Step 2: Reverse each row        for (int i = 0; i < n; i++) {            for (int j = 0; j < n / 2; j++) {                int temp = matrix[i][j];                matrix[i][j] = matrix[i][n - 1 - j];                matrix[i][n - 1 - j] = temp;            }        }    }}