Problem Description

The problem presents a scenario where you start with an initial integer array arr, and each day a new array is derived from the array of the previous day through a set of specific transformations. The rules for transforming any given array element are:

1. If an element is smaller than both its immediate neighbors (left and right), increase its value by 1.
2. If an element is larger than both its immediate neighbors, decrease its value by 1.
3. The first and last elements of the array remain unchanged regardless of their neighbors.

This process is to be repeated day after day until eventually, the array reaches a state where no further changes occur - in other words, the array becomes stable and does not change from one day to the next. The objective is to determine what this final array looks like.

Intuition

To solve this problem, we can simulate the array transformation process day by day until the array becomes stable. Here’s how we can arrive at the solution approach:

1. Create a flag (f) to keep track of whether any change has occurred on a given day.
2. Iterate over the array, starting from the second element and ending at the second-to-last element (since the first and last elements do not change).
3. For each element, compare it with its left and right neighbors to decide whether to increment or decrement it according to the rules.
   • If the current element is smaller than both neighbors, increment it.
   • If the current element is larger than both neighbors, decrement it.
4. To avoid impacting the comparison of subsequent elements, store the original array in a temporary array (t) before making any changes.
5. Continue the day-by-day simulation until a day passes where no changes are made to the array, indicating the array has become stable.
6. When no changes occur (the flag f remains false after a full iteration), return the stable array.

The process relies on careful iteration and conditionally applying the transformation rules, keeping an unchanged reference of the array to determine whether the current element needs to be changed without affecting subsequent comparisons.

Solution Approach

The solution approach follows a straightforward brute-force method to simulate the day-to-day transformation of the array until it reaches a state of equilibrium where no further changes occur. Here's how the implementation of the solution is carried out:

1. The solution uses a while loop flagged by a boolean f which is initially set to True. This flag f is used to check whether any changes are made to the array in the current iteration (day). If no changes are made, f remains False and the loop ends.

2. At the start of each iteration, the current array (arr) is cloned into a temporary array (t). This is important because we want to evaluate the elements against their original neighbors, and modifying arr in-place would disturb those comparisons.

3. The implementation then iterates over all the elements of the array starting from index 1 and ending at len(arr) - 2 to ensure the first and last elements remain unchanged, in compliance with rule 3 of the problem description.

4. During each iteration, the algorithm checks each element against its neighbors:

   • If t[i] (the element in the temporary array) is greater than both t[i - 1] and t[i + 1], the element in the original array arr[i] is decremented by 1, and f is set to True indicating a change.

   if t[i] > t[i - 1] and t[i] > t[i + 1]:
       arr[i] -= 1
       f = True

   • If t[i] is less than both t[i - 1] and t[i + 1], arr[i] is incremented by 1, and again, f is set to True.

   if t[i] < t[i - 1] and t[i] < t[i + 1]:
       arr[i] += 1
       f = True

5. If f is False after the inner loop, it means no changes were made in the latest day, so the array has reached its final stable state and the loop exits.

6. The final stable array (arr) is then returned.

The solution does not use any complex algorithms or data structures; it is a simple iterative approach that exploits array indexing and conditional logic to simulate the situation described in the problem. It ensures the integrity of comparisons through the use of a temporary array and signals the completion of the simulation using a loop control variable.

Example Walkthrough

Let's say we have an initial integer array arr = [6, 4, 8, 2, 3]. We want to apply the solution approach to this array to find out what the final stable array will look like after applying the transformations day after day as per the rules stated.

Day 1:

• We create a temporary array t which is a copy of arr: t = [6, 4, 8, 2, 3].
• We iterate starting from the second element to the second to last:
  • t[1] = 4, it's less than both its neighbors t[0] = 6 and t[2] = 8, so arr[1] becomes 4 + 1 = 5.
  • t[2] = 8, it's greater than both t[1] = 4 (before increment) and t[3] = 2, so arr[2] becomes 8 - 1 = 7.
  • t[3] = 2, it's less than t[2] = 8 (before decrement) and t[4] = 3, so arr[3] becomes 2 + 1 = 3.
• The array arr at the end of Day 1 is [6, 5, 7, 3, 3].

Day 2:

• We clone the array again: t = [6, 5, 7, 3, 3].
• Iterating through t:
  • t[1] = 5, no change as it is not less than or greater than both neighbors.
  • t[2] = 7, no change as it is not less than or greater than both neighbors.
  • t[3] = 3, no change as it is not less than or greater than both neighbors.
• The array arr at the end of Day 2 is [6, 5, 7, 3, 3].

No elements in arr changed during Day 2, so now we know that the array has become stable, and the process can end here. The final array is [6, 5, 7, 3, 3]. This array will remain unchanged in subsequent days, as it meets none of the conditions for incrementing or decrementing any of its elements (except for the first and last elements, which do not change anyway).

Therefore, our final stable array, after applying the given transformation rules, is [6, 5, 7, 3, 3].

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is primarily dependent on two nested loops: the outer while loop and the inner for loop.

• The outer while loop continues executing until no more changes are made to the array. In the worst case scenario, it could run for a number of iterations proportional to the size of the array, n. This is because each iteration could potentially only decrease or increase each element by 1, and for the array to become stable, it might require multiple single-unit adjustments at different positions.
• The inner for loop goes through the elements of the array, starting from 1 to len(arr) - 2. This for loop has a time complexity of O(n - 2), which simplifies to O(n).

Therefore, in the worst case, the time complexity of the entire algorithm becomes O(n^2) since for each iteration of the while loop, a full pass through most of the array is completed using the for loop.

Space Complexity

The space complexity of the code consists of:

• A temporary array t that is a copy of the input array arr. This copy is made in each iteration of the while loop. The temporary array t has the same size as the input array, which gives us a space complexity of O(n).
• No other significant extra space is used, as the operations are performed in place with just a few extra variables (f, i) whose space usage is negligible.

So, the total space complexity of the algorithm is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.