Max Consecutive Ones III
<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>
Array; Binary Search; Prefix Sum; Sliding Window
class Solution {
public:
    int longestOnes(vector<int>& nums, int k) {
        int l = 0, cnt = 0;
        for (int x : nums) {
            cnt += x ^ 1;
            if (cnt > k) {
                cnt -= nums[l++] ^ 1;
            }
        }
        return nums.size() - l;
    }
};