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# 1.7 Longest Subarray with Ones after Replacement (hard)

**Problem:** 

[Max Consecutive Ones III](https://leetcode.com/problems/max-consecutive-ones-iii/)

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.

**Example 1:**

```
Input: nums = [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: nums = [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.
```

**Solution:**

```java
public int longestOnes(int[] A, int K) {
    int max = 0;
    int zeroCount = 0; // zero count in current window
    int i = 0; // slow pointer
    for(int j = 0; j < A.length; ++j) {
        if(A[j] == 0) { // move forward j, if current is 0, increase the zeroCount
            zeroCount++;
        }
        
        // when current window has more than K, the window is not valid any more
        // we need to loop the slow pointer until the current window is valid
        while(zeroCount > K) {  
            if(A[i] == 0) {
                zeroCount--;
            }
            i++;
        }
        max = Math.max(max, j-i+1); // everytime we get here, the current window is valid 
    }
    return max;
}
```


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