## Description

Given an array A of 0s and 1s, we may change up to K values from 0 to 1. Return the length of the longest (contiguous) subarray that contains only 1s.

Example 1: Input: A = [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: A = [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.

Note: 1 <= A.length <= 20000 0 <= K <= A.length A[i] is 0 or 1

## Solution

We can think this problem in another way. Lets use classic sliding window template in the thread

## Code

```
class Solution {
public:
int longestOnes(vector<int>& A, int K) {
int maxLength = 0;
int left = 0;
int count = 0;
for(int i = 0; i < A.size(); ++i) {
if(A[i] == 0) { /*❶*/
count++;
}
while(count > K) { /*❷*/
if(A[left] == 0){
count--;
}
left++;
}
maxLength = max(maxLength, i - left + 1);/*❸*/
}
return maxLength;
}
};
```

❶ update count, it is related to K.

❷ move left bound of window and update the count.

❸ update the length of window.

## Insight

There are lots of questions can be solved by similar way. And sliding window technique is very popular approach.