Minimum Operations for Stepwise Structures

You are given an integer array structures, where structures[i] is the height of the structure at position i.
In one operation, you may increase the height of any one structure by exactly 1. You may perform as many operations as needed, but you may not decrease any height.
Transform the array into either of these stepwise patterns:
Ascending: every structure is exactly 1 unit taller than the structure immediately before it.
Descending: every structure is exactly 1 unit shorter than the structure immediately before it.
Return the minimum number of operations needed to obtain either pattern.

Function
minimumStepwiseOperations(structures: int[]) → long

Examples
Example 1
structures = [1,4,3,2]
return = 4
Add 4 units to the first structure. The final heights are [5,4,3,2], which form a descending stepwise pattern.
Example 2
structures = [5,7,9,4,11]
return = 9
Add 2 units to the first structure, 1 unit to the second, and 6 units to the fourth. The final heights are [7,8,9,10,11], which form an ascending stepwise pattern, using 9 operations.

Constraints
structures is an array of integers representing structure heights.
An operation increases exactly one structure height by 1.
The final adjacent heights must differ by exactly 1 throughout the array.
The source shows a Java execution time limit of 3 seconds and a memory limit of 1 GB.

Here's the reduction. For an ascending target, the final height at index i is base + i. You can only increase, so base + i >= structures[i] for every i. The smallest valid base is max(structures[i] - i). Cost is the sum of (base + i - structures[i]). For descending, the target is base - i, so you need base >= structures[i] + i. Take base = max(structures[i] + i) and sum the gaps. Return the smaller of the two totals. It's O(n) with two passes. The pitfall is overflow, since the return type is long, so accumulate in a long. Another trap is trying to pick a base by simulation or binary search. The max formula is exact, because a larger base only adds cost. StealthCoder is the hedge if the formula slips your mind live.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.