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Tested Python solution for LeetCode 832 with 17 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 832 # by problem number
lcpy gen -s flipping_an_image # by problem name
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.
[1,1,0] horizontally results in [0,1,1].0 is replaced by 1, and each 1 is replaced by 0.
[0,1,1] results in [1,0,0].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]]
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]]
n == image.lengthn == image[i].length1 <= n <= 20images[i][j] is either 0 or 1.class Solution:
# Time: O(n^2)
# Space: O(1) extra (output not counted)
def flip_and_invert_image(self, image: list[list[int]]) -> list[list[int]]:
for row in image:
left, right = 0, len(row) - 1
while left < right:
row[left], row[right] = 1 - row[right], 1 - row[left]
left += 1
right -= 1
if left == right:
row[left] = 1 - row[left]
return image
| Time | Space |
|---|---|
| O(n^2) | O(1) extra (output not counted) |