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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.

 

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]]

 

Constraints:

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

Solution Explanation: Rotating an Image In-Place

This problem challenges us to rotate a square matrix (image) 90 degrees clockwise in-place, meaning without using extra space proportional to the input size. The optimal approach leverages two key steps:

1. Transpose: A matrix transpose swaps rows and columns. If we transpose the input matrix, we're halfway to a 90-degree rotation.

2. Reverse Rows: After the transpose, we need to reverse each row to complete the 90-degree clockwise rotation.

Why this works: Consider a single element at matrix[i][j].

  • Transpose: The element moves to matrix[j][i].
  • Reverse Rows: In the j-th row, the element at index i moves to index n-1-i, where n is the dimension of the square matrix. Therefore, the final position of the element is matrix[j][n-1-i], which is precisely the location after a 90-degree clockwise rotation.

Time Complexity: O(n²) - We iterate through the matrix twice (transpose and row reversal), each iteration taking O(n²) time.

Space Complexity: O(1) - We perform operations directly on the input matrix, using only a constant amount of extra space.

Code Implementations

The following code snippets demonstrate the solution across various programming languages. They all follow the same two-step approach: transpose and reverse rows. Note the slight variations in how the transpose and row reversal are implemented for optimal efficiency.

Python:

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        n = len(matrix)
        # Transpose
        for i in range(n):
            for j in range(i, n):
                matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
 
        # Reverse rows
        for i in range(n):
            matrix[i].reverse()

Java:

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        // Transpose
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
 
        // Reverse rows
        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;
            }
        }
    }
}

C++:

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        // Transpose
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                swap(matrix[i][j], matrix[j][i]);
            }
        }
 
        // Reverse rows
        for (int i = 0; i < n; i++) {
            reverse(matrix[i].begin(), matrix[i].end());
        }
    }
};

JavaScript:

var rotate = function(matrix) {
    let n = matrix.length;
    // Transpose
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
        }
    }
 
    // Reverse rows
    for (let i = 0; i < n; i++) {
        matrix[i].reverse();
    }
};

These are examples; other languages would follow a similar structure. The core idea remains the same: efficiently transpose and then reverse rows to achieve the in-place 90-degree clockwise rotation.