LeetCode 1243. Array Transformation Solution in Java, C++, Python & Go | Explanation + Code

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1243. Array Transformation

Description

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:

  1. If an element is smaller than both its left neighbor and its right neighbor, then this element is incremented.
  2. If an element is bigger than both its left neighbor and its right neighbor, then this element is decremented.
  3. 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

Solutions

Solution 1: Simulation

Simulate each day. For each element, if it is greater than its left and right neighbors, it decreases by 1, otherwise, it increases by 1. If the array no longer changes on a certain day, return that array.

The time complexity is O(n × m), and the space complexity is O(n). Where n is the length of the array, and m is the maximum value in the array.

PythonJavaC++Go
class Solution: def transformArray(self, arr: List[int]) -> List[int]: f = True while f: f = False t = arr[:] for i in range(1, len(t) - 1): if t[i] > t[i - 1] and t[i] > t[i + 1]: arr[i] -= 1 f = True if t[i] < t[i - 1] and t[i] < t[i + 1]: arr[i] += 1 f = True return arr(code-box)

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