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53 lines
1.5 KiB
Plaintext
53 lines
1.5 KiB
Plaintext
---
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title: '1004. Max Consecutive Ones III'
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description: 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
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icon: dot
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topics:
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- { name: "Array", slug: "array" }
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- { name: "Binary Search", slug: "binary-search" }
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- { name: "Sliding Window", slug: "sliding-window" }
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- { name: "Prefix Sum", slug: "prefix-sum" }
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issue: "https://github.com/prdlk/leetcode/issues/56"
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sidebar:
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label: 'Max Consecutive Ones III'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2`
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- Output: `6`
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- 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.
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### Example 2:
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- Input: `nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3`
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- Output: `10`
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- 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.
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### Constraints:
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- `1 <= nums.length <= 10^5`
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- `nums[i]` is either 0 or 1.
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- `0 <= k <= nums.length`
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## Solution
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```py
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class Solution:
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def longestOnes(self, nums: list[int], k: int) -> int:
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left = 0
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cur = 0
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ans = 0
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for right in range(len(nums)):
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if nums[right] == 0:
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cur += 1
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while cur > k:
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if nums[left] == 0:
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cur -= 1
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left += 1
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ans = max(ans, right - left + 1)
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return ans
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```
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