1473. Paint House III

Problem Link

Description

---

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 have 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 x n matrix cost and an integer target where:

• houses[i]: is the color of the house i, and 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 it is 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 = [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] <= 104

Solution

---

Python3

class Solution:
    def minCost(self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int) -> int:
        INF = float('inf')
        
        @cache
        def go(i, prev, k):
            if k > target: return INF
            
            if i == m:
                return 0 if k == target else INF
            
            if houses[i] != 0:
                if houses[i] == prev:
                    return go(i + 1, houses[i], k)
                else:
                    return go(i + 1, houses[i], k + 1)
            
            res = INF
            
            for j, paintCost in enumerate(cost[i]):
                if j + 1 == prev:
                    res = min(res, paintCost + go(i + 1, j + 1, k))
                else:
                    res = min(res, paintCost + go(i + 1, j + 1, k + 1))
            
            return res
        
        ans = go(0, -1, 0)
        return ans if ans != INF else -1