# LeetCode 832. Flipping an Image

## Description

https://leetcode.com/problems/flipping-an-image/

Given an `n x n` binary matrix `image`, flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed.

• For example, flipping `[1,1,0]` horizontally results in `[0,1,1]`.

To invert an image means that each `0` is replaced by `1`, and each `1` is replaced by `0`.

• For example, inverting `[0,1,1]` results in `[1,0,0]`.

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`.

## Explanation

insert into a new matrix base on problem description.

## Python Solution

``````class Solution:
def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
flip_image = [[] for i in range(len(image))]

for i in range(len(image)):
for j in range(len(image[0])):

flip_image[i].append(0 if image[i][len(image[0]) - 1 - j] == 1 else 1)

return flip_image``````
• Time Complexity: O(N^2)
• Space Complexity: O(N^2)