[leetcode] 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]]

 

 

 

Solution

from typing import List
import copy

input1 = [[1,1,0],[1,0,1],[0,0,0]]
# output1 = [[1,0,0],[0,1,0],[1,1,1]]
input2 = [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
# output2 = [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]

def flipAndInvertImage(A: List[List[int]]) -> List[List[int]]:
    
    # (0) copy the input
    B = copy.deepcopy(A)
    #B = A[:] # shallow copy
    
    # (1) horizontally flipping
    for i in range(len(B)):
        for j in range(len(B[0])):
            if j < len(B[0])/2:
                B[i][j], B[i][-1-j] = B[i][-1-j], B[i][j]
            else:
                break
    # (2) inverting
    for i in range(len(A)):
        for j in range(len(A[0])):
            if B[i][j] == 1:
                B[i][j] = 0
            else: # B[i][j] == 0
                B[i][j] = 1     
    
    # (3) return
    return B

 

 

 

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