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).