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)

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