<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 {
public:
    int longestOnes(vector<int>& nums, int k) {
        vector<int> fi(k);
        int flip = k;
        int j = 0;
        int ans = 0, cur = 0;
        if(k == 0){
            for(int i = 0;i<nums.size();i++){
                if(nums[i] == 1) cur++;
                else{
                    ans = max(ans, cur);
                    cur = 0;
                }
            }
            return max(ans,cur);
        }
        for(int i = 0;i<nums.size();i++){
            if(nums[i] == 1) cur++;
            else{
                if(flip){
                    flip--;
                    cur++;
                    fi[j] = i;
                    j++;
                }
                else{
                    cur = i - fi[j%k];
                    fi[j%k] = i;
                    j++;
                }
            }
            ans = max(ans, cur);
        }
        return ans;
    }
};
59000