48. Rotate Image

Problem #48 (Rotate Image | Array, Math, Matrix)

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.

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Example 1

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

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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]]

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Constraints

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

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SOLUTIONS

Matrix => Transpose => Horizontal Flip(In-place)

This solution is done by getting the transpose of the 2D Matrix and flipping each of the rows horizontally.

Matrix:

Transpose:

• The transpose of the matrix is basically done by swapping the values of matrix[i][j] and matrix[i][j].

Horizontal Flip:

• Flip the matrix horizontally by swapping the values of matrix[i][j] and matrix[i][matrix.length - 1 - j], where j < matrix.length/2.

CODE <br.>

• JAVA

class Solution {
    public void rotate(int[][] matrix) {
        for(int i = 0; i < matrix.length; i++){
            for(int j = i; j < matrix[0].length; j++){
                int t = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = t;
            }
        }
        
        for(int i = 0; i < matrix.length; i++){
            for(int j = 0; j < matrix.length/2; j++){
                int t = matrix[i][j];
                matrix[i][j] = matrix[i][matrix.length - 1 - j];
                matrix[i][matrix.length - 1 - j] = t;
            }
        }
    }
}

• C++

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        for(int i = 0; i < matrix.size(); i++){
            for(int j = i; j < matrix[0].size(); j++){
                int t = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = t;
            }
        }
        
        for(int i = 0; i < matrix.size(); i++){
            for(int j = 0; j < matrix.size()/2; j++){
                int t = matrix[i][j];
                matrix[i][j] = matrix[i][matrix.size() - 1 - j];
                matrix[i][matrix.size() - 1 - j] = t;
            }
        }
    }
};

Complexity:

• Time: O(m * n), where m is the number of rows and n is the number of columns.
• Spaca: O(1)