Flipping an Image

Problem Statement

In the provided problem, we are given a binary matrix image where n x n denotes its dimensions. The task is to manipulate this matrix by performing two specific transformations:

1. Flip the matrix horizontally. This means reversing the order of elements in each row of the matrix. For instance, if a row in the matrix is [1,1,0], after a horizontal flip it would become [0,1,1].
2. Invert the matrix. Post the horizontal flip; each element in the matrix needs to be inverted. In this context, inversion means converting every 1 to 0 and every 0 to 1. For example, after flipping if a row reads [0,1,1], it should be transformed to [1,0,0] after the inversion.

The end goal is to return the matrix after these transformations have been applied to each row.

Examples

Example 1

Input:

image = [[1,1,0],[1,0,1],[0,0,0]]

Output:

[[1,0,0],[0,1,0],[1,1,1]]

Explanation:

First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]

Example 2

Input:

image = [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]

Output:

[[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]

Explanation:

First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]

Constraints

• n == image.length
• n == image[i].length
• 1 <= n <= 20
• images[i][j] is either 0 or 1.

Approach and Intuition

The process of transforming the matrix can be understood and implemented through the following steps:

1. Flip Horizontally:

   • Traverse through each row of the matrix.
   • Reverse the order of elements within the row. This can be done using slicing in Python (row[::-1]), or any similar method in other programming languages which supports reversing lists or arrays.

2. Invert the Matrix:

   • After flipping all the rows horizontally, iterate through each element of each row.
   • Replace 1 with 0 and 0 with 1. This can involve a conditional check for each element (e.g., if element is 1 then it becomes 0).

Given these operations and based on our constraints whereby the dimensions and values are within a manageable range (n up to 20 and matrix elements as 0 or 1), an approach that involves iterating through each element in the matrix sequentially to perform the required transformations is computationally feasible.

Thus, the algorithm involves running two nested loops: the outer loop iterates through each row and the inner loop iterates through each element of the row, ensuring all elements are first reversed and then inverted. The entire process runs in O(n^2) time complexity where n is the dimension of the matrix, making it a straightforward and efficient solution for the given problem size constraints.

Solutions

class Solution {
    public int[][] invertAndFlipImage(int[][] image) {
        int cols = image[0].length;
        for (int[] row: image)
            for (int j = 0; j < (cols + 1) / 2; ++j) {
                int temp = row[j] ^ 1;
                row[j] = row[cols - 1 - j] ^ 1;
                row[cols - 1 - j] = temp;
            }

        return image;
    }
}