Skip to content

861. Flipping An Image

Array Two Pointers Bit Manipulation Matrix Simulation

Problem - Flipping An Image

Easy

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.

Solutions

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        n = len(image)
        for row in image:
            i, j = 0, n - 1
            while j > i:
                if row[i] == row[j]:
                    row[i] ^= 1
                    row[j] ^= 1
                i, j = i + 1, j - 1
            if i == j:
                row[i] ^= 1
        return image

Submission Stats:

  • Runtime: 0 ms (100.00%)
  • Memory: 17.7 MB (89.04%)