<p>Given a binary array <code>nums</code> and an integer <code>k</code>, return <em>the maximum number of consecutive </em><code>1</code><em>&#39;s in the array if you can flip at most</em> <code>k</code> <code>0</code>&#39;s.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> [1,1,1,0,0,<u><strong>1</strong>,1,1,1,1,<strong>1</strong></u>]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
	<li><code>0 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        l = 0
        zeros = 0
        ans = 0
        
        for r in range(len(nums)):
            if nums[r] == 0:
                zeros += 1
            while zeros > k:
                if nums[l] == 0:
                    zeros -= 1
                l += 1
            ans = max(ans, r - l + 1)
            r += 1
        return ans
16500