## 1. Problem

1493. Longest Subarray of 1’s After Deleting One Element

Given a binary array nums, you should delete one element from it.

Return the size of the longest non-empty subarray containing only 1’s in the resulting array. Return 0 if there is no such subarray.

**Example 1:**

```
Input: nums = [1,1,0,1]
Output: 3
Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.
```

**Example 2:**

```
Input: nums = [0,1,1,1,0,1,1,0,1]
Output: 5
Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].
```

**Example 3:**

```
Input: nums = [1,1,1]
Output: 2
Explanation: You must delete one element.
```

**Constraints:**

- 1 <= nums.length <= 10^5
- nums[i] is either 0 or 1.

## 2. Solution

I solved this problem like this.

- Solve the problem with sliding window!

### 2.1. Solution1

- Check used zero count. If zero count is more bigger than k(max zero count),

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

### 2.2. Solution2

- Minus k value, if now idx value is 0.
- if k is smaller than 0, move left and plus k value if left is 0.
- return r - l. Once the maximum length is determined, the r - l value is also equal to the maximum length.

```
class Solution:
def longestSubarray(self, nums: List[int]) -> int:
k = 1
l = 0
for r in range(len(nums)):
if nums[r] == 0:
k-=1
if k < 0:
if nums[l] == 0:
k += 1
l += 1
return r - l
```