[CODE ONLY] PYTHON USING Ajay Prem's SOLUTION
Using Ajay Prem's solution:LINK: https://leetcode.com/problems/paint-house-iv/solutions/6301141/fold-cost-in-half-dp-pythonBelow is my solution that was refacto
# Using Ajay Prem's solution:
# LINK: https://leetcode.com/problems/paint-house-iv/solutions/6301141/fold-cost-in-half-dp-python


# Below is my solution that was refactored for explicit clarity...

# Code
```python3 []
# Using Ajay Prem's solution:
# LINK: https://leetcode.com/problems/paint-house-iv/solutions/6301141/fold-cost-in-half-dp-python

class Solution:
    def minCost(self, n: int, cost: List[List[int]]) -> int:
        condensed_cost = []

        half_array = n // 2

        for i in range(half_array):
            first_half_array_cost = cost[i]
            second_half_array_cost = cost[n - i - 1]
            condensed_pair = [first_half_array_cost, second_half_array_cost]
            condensed_cost.append(condensed_pair)

        half_n = n // 2

        @cache
        def helper(house_index, last_chosen_first, last_chosen_second):
            if house_index == half_n:
                return 0

            best_choice = inf
            for first_choice in range(3):
                for second_choice in range(3):
                    if first_choice == last_chosen_first:
                        continue
                    if second_choice == last_chosen_second:
                        continue
                    if first_choice == second_choice:
                        continue
                    cost_first = condensed_cost[house_index][0][first_choice]
                    cost_second = condensed_cost[house_index][1][second_choice]
                    neighbor_house_cost = helper(house_index + 1, first_choice, second_choice)
                    subtotal = cost_first + cost_second + neighbor_house_cost
                    best_choice = min(best_choice, subtotal)
            return best_choice
        return helper(0, -1, -1)


            
```
['Python3']