Max Consecutive Ones 3
Given an array A of 0s and 1s, we may change up to K values from 0 to 1.
Return the length of the longest (contiguous) subarray that contains only 1s.
Example 1:
Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
Output: 6
Explanation:
[1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.class Solution:
def longestOnes(self, A: List[int], K: int) -> int:
self.answer = 0
start, end, zeros, length = 0, 0, 0, 0
while end < len(A):
if A[end] == 0:
zeros += 1
while zeros > K:
if A[start] == 0:
zeros -= 1
length -= 1
start += 1
length += 1
self.answer = length if length > self.answer else self.answer
end += 1
return self.answerLast updated
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