class Solution:
def flipAndInvertImage(self, A: List[List[int]]) -> List[List[int]]:
for i in range(len(A)):
A[i].reverse()
for i in range(len(A)):
for j in range(len(A[i])):
if A[i][j] ==0:
A[i][j] = 1
else:
A[i][j] = 0
return A
2148. Count Elements With Strictly Smaller and Greater Elements | 2149. Rearrange Array Elements by Sign |
2150. Find All Lonely Numbers in the Array | 2151. Maximum Good People Based on Statements |
2144. Minimum Cost of Buying Candies With Discount | Non empty subsets |
1630A - And Matching | 1630B - Range and Partition |
1630C - Paint the Middle | 1630D - Flipping Range |
1328A - Divisibility Problem | 339A - Helpful Maths |
4A - Watermelon | 476A - Dreamoon and Stairs |
1409A - Yet Another Two Integers Problem | 977A - Wrong Subtraction |
263A - Beautiful Matrix | 180C - Letter |
151A - Soft Drinking | 1352A - Sum of Round Numbers |
281A - Word Capitalization | 1646A - Square Counting |
266A - Stones on the Table | 61A - Ultra-Fast Mathematician |
148A - Insomnia cure | 1650A - Deletions of Two Adjacent Letters |
1512A - Spy Detected | 282A - Bit++ |
69A - Young Physicist | 1651A - Playoff |