Max consecutive ones iii

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        l = r = 0
        flips = k
        res = 0
        while r < len(nums):
            if nums[r] == 0:
                flips -= 1
            if flips < 0:
                if nums[l] == 0:
                    flips += 1
                l += 1
            r += 1
            res = max(res, r - l)
        return res

Max Consecutive Ones III

---

Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

 

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.
• 0 <= k <= nums.length