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

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

def _disabled_input(*args, **kwargs):
    raise RuntimeError("input() is disabled in callable mode")
builtins.input = _disabled_input

ns = {"__name__": "__judge__"}  # prevent if __name__ == "__main__" blocks
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "# https://zxi.mytechroad.com/blog/dynamic-programming/leetcode-1473-paint-house-iii/\n# k: num of neighborhoods, i: num of houses, c: color of the last house\n# dp[k][i][c]: min cost to form k neighborhoods using the first i houses and the i-th house's color is c\n# init: dp[0][0][*] = 0, else is inf\n# if the i - 1 house's color != i house's color, means they are not in the same neighborhood\n# dp[k][i][ci] = dp[k - 1][i - 1][ci-1] + cost\n# else if the i - 1house's color == i house's color, means they are in the same neighborhood\n# dp[k][i][ci] = dp[k][i - 1][ci-1] + cost\n# if houses[i] == ci, house i is painted by color i, cost = 0, no need to paint\n# else, cost = cost[i][ci]\n# if ci != houses[i], means house i is painted by another color, dp[k][i][ci] = inf\n# ans = min(dp[target][m][*])\n# 因为需要枚举所有的 house i - 1 和 house i 的颜色的组合, 所以时间复杂度乘以 n * n\n# 因为 dp[k] 只和 dp[k] 和 dp[k - 1] 有关, 所以时间复杂度可以从 O(target * m * n) -> O(m * n)\n# O(target * m * n * n) time compleixty, O(m * n) space complexity\nclass Solution:\n    def minCost(self, houses: List[int], costs: List[List[int]], m: int, n: int, target: int) -> int:\n        dp = [[[float('inf') for _ in range(n + 1)] for _ in range(m + 1)] for _ in range(target + 1)]\n        for c in range(n + 1):\n            dp[0][0][c] = 0\n\n        for k in range(1, target + 1):\n            for i in range(k, m + 1):  # 因为 i < k 的情况不合法, 组成 k 个 neighborhoods 至少也需要 k 个 house\n                hi = houses[i - 1]\n                hj = 0  # 初始化前一个房子的颜色为 0, 如果 i < 2, 则它就是第一个房子, 所以设前一个房子可以是任何颜色的, 即 0\n                if i >= 2:\n                    hj = houses[i - 2]\n                for ci in range(1, n + 1):\n                    if hi != 0 and hi != ci:  # 当这个房子 i 已经有颜色且不是当前想给它的颜色时, 直接跳过\n                        dp[k][i][ci] = float('inf')\n                        continue\n                    cost = 0  # 如果这个房子已经有颜色且和当前想给它的颜色一样, cost 为 0\n                    if hi != ci:  # 如果这个房子没有颜色, cost 为 costs 矩阵中的值\n                        cost = costs[i - 1][ci - 1]\n                    for cj in range(1, n + 1):\n                        dp[k][i][ci] = min(dp[k][i][ci], dp[k - (ci != cj)][i - 1][cj] + cost)\n        \n        res = min(dp[target][m][c] for c in range(1, n + 1))\n        if res == float('inf'):\n            return -1\n        return res\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'minCost'
if _fn_name:
    if "Solution" in ns:
        try:
            _obj = ns["Solution"]()
            if hasattr(_obj, _fn_name):
                _target = getattr(_obj, _fn_name)
        except Exception:
            _target = None
    if _target is None and _fn_name in ns and callable(ns[_fn_name]):
        _target = ns[_fn_name]
# Fallback: if there's exactly one user-defined function, use it
if _target is None:
    _cands = [v for k, v in ns.items() if callable(v) and getattr(v, "__module__", "") in ("__judge__", "__main__")]
    _user_funcs = [c for c in _cands if getattr(c, "__name__", "").startswith("_") is False]
    if len(_user_funcs) == 1:
        _target = _user_funcs[0]
if _target is None:
    raise RuntimeError(f"Could not resolve callable '{_fn_name}'")

_raw = sys.stdin.read().strip()
if _raw:
    _parsed = json.loads(_raw)
    if isinstance(_parsed, dict):
        _kwargs = _parsed.get("kwargs", {})
        _args = _parsed.get("args", [])
    else:
        _args, _kwargs = _parsed, {}
else:
    _args, _kwargs = [], {}
if not isinstance(_args, (list, tuple)):
    _args = [_args]
_res = _target(*_args, **_kwargs)
try:
    sys.stdout.write(repr(_res) + "\n")
except Exception:
    pass

```

Measured memory_bytes: 11724