human
I found an interesting problem on https://leetcode.com/problems/paint-house-iii/:
There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that has been painted last summer should not be painted again.
A neighborhood is a maximal group of continuous houses that are painted with the same color. (For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods  [{1}, {2,2}, {3,3}, {2}, {1,1}]).
Given an array houses, an m * n matrix cost and an integer target where:

houses[i]: is the color of the house i, 0 if the house is not painted yet.
cost[i][j]: is the cost of paint the house i with the color j+1.

Return the minimum cost of painting all the remaining houses in such a way that there are exactly target neighborhoods, if not possible return -1.
 
Example 1:
Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 9
Explanation: Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.

Example 2:
Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 11
Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. 
Cost of paint the first and last house (10 + 1) = 11.

Example 3:
Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[1,10],[10,1],[1,10]], m = 5, n = 2, target = 5
Output: 5

Example 4:
Input: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
Output: -1
Explanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.

 
Constraints:

m == houses.length == cost.length
n == cost[i].length
1 <= m <= 100
1 <= n <= 20
1 <= target <= m
0 <= houses[i] <= n
1 <= cost[i][j] <= 10^4

I tried it in Python, but could not do it. Can you solve it?

 ... Give me a the Python program you requested?
assistant
#Here is the Python program you requested.

from collections import defaultdict
class Solution:
    def minCost(self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int) -> int:
        '''
        
        m houses - n colors (1~n)
        neighborhoods of same color
        0 = not colored yet
        cost[i][j] - cost to paint i with color j+1
        
        return minimal cost of painting remaining houses to get exactly target neighborhoods
        
        -have to keep track of min cost
        -num neighbors==target at end
        
        dp => cost,numNeighbors
        dp[house][lastneighbor] = min(dp[house][lastneighbor], min(cost + dp[house-1][neighbors])
        
        best now = last house with same color, correct num of neighbors or last house diff color, 1 less num of neighbors
        
        dp[h][numNeighbors][color] = colorCost + min(
                dp[h-1][numNeighbors][color],
                dp[h-1][numNeighbors-1][diffColor] <- iterate
            )
        
        at house h,
        dp[neighbors][color] = colorCost (if it's 0)
                                + prev[neighbors][color]
                                + prev[neighbors-1][anothercolor]
        neighbors: [j-target,j+1]
        
        edge cases:
        -if # of color neighborhoods > target: return -1
        -num houses < target: return -1
        '''
        prev = [[0]*n for _ in range(target+1)]
        for h,house in enumerate(houses):
            dp = [[float('inf')]*n for _ in range(target+1)]
            for numNeighbors in range(1,min(h+2,target+1)):
                if house==0:
                    for color in range(n):
                        colorCost = cost[h][color]
                        dp[numNeighbors][color] = min(
                            dp[numNeighbors][color],
                            colorCost + prev[numNeighbors][color],
                            colorCost + min(prev[numNeighbors-1][c] for c in range(n) if c!=color)
                        )
                else:
                    color = house-1
       ...