## Description

https://leetcode.com/problems/array-transformation/

Given an initial array `arr`

, every day you produce a new array using the array of the previous day.

On the `i`

-th day, you do the following operations on the array of day `i-1`

to produce the array of day `i`

:

- If an element is smaller than both its left neighbor and its right neighbor, then this element is incremented.
- If an element is bigger than both its left neighbor and its right neighbor, then this element is decremented.
- The first and last elements never change.

After some days, the array does not change. Return that final array.

**Example 1:**

Input:arr = [6,2,3,4]Output:[6,3,3,4]Explanation:On the first day, the array is changed from [6,2,3,4] to [6,3,3,4]. No more operations can be done to this array.

**Example 2:**

Input:arr = [1,6,3,4,3,5]Output:[1,4,4,4,4,5]Explanation:On the first day, the array is changed from [1,6,3,4,3,5] to [1,5,4,3,4,5]. On the second day, the array is changed from [1,5,4,3,4,5] to [1,4,4,4,4,5]. No more operations can be done to this array.

**Constraints:**

`3 <= arr.length <= 100`

`1 <= arr[i] <= 100`

## Explanation

The list length indicates how many times we need to transform the array.

## Python Solution

```
class Solution:
def transformArray(self, arr: List[int]) -> List[int]:
arr_num = list(arr)
for k in range(len(arr_num)):
results = []
for i, num in enumerate(arr_num):
if i != 0 and i != len(arr_num) - 1:
if num > arr_num[i - 1] and num > arr_num[i + 1]:
results.append(num - 1)
elif num < arr_num[i - 1] and num < arr_num[i + 1]:
results.append(num + 1)
else:
results.append(num)
else:
results.append(num)
arr_num = results
return arr_num
```

- Time Complexity: O(N^2).
- Space Complexity: O(N).