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
class Solution:
    def minCost(self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int) -> int:
        A=houses
        dp, dp2 = {(0, 0): 0}, {}
        for i, a in enumerate(A):
            for cj in (range(1, n + 1) if a == 0 else [a]):
                for ci, b in dp:
                    b2 = b + (ci != cj)
                    if b2 > target: continue
                    dp2[cj, b2] = min(dp2.get((cj,b2), float('inf')), dp[ci, b] + (cost[i][cj - 1] if cj != a else 0))
            dp, dp2 = dp2, {}
        return min([dp[c, b] for c, b in dp if b == target] or [-1])
class Solution:
    def minCost(self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int) -> int:
        A=houses
        dp, dp2 = {(0, 0): 0}, {}
        for i, a in enumerate(A):
            for cj in (range(1, n + 1) if a == 0 else [a]):
                for ci, b in dp:
                    b2 = b + (ci != cj)
                    if b2 > target: continue
                    dp2[cj, b2] = min(dp2.get((cj,b2), float('inf')), dp[ci, b] + (cost[i][cj - 1] if cj != a else 0))
            dp, dp2 = dp2, {}
        return min([dp[c, b] for c, b in dp if b == target] or [-1])