1473. Paint House III

Hard

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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 : Took help from youtube to solve this

class Solution:
    def minCost(self, houses: List[int], cost: List[List[int]], m: int, n: int, target: int) -> int:
        dp = {}
        
        def dfs(index, target, prev_color):
            key = (index, target, prev_color)
            if index == len(houses) or target < 0 or (m - index) < target:
                return 0 if target == 0 and index == len(houses) else float("inf")
            if key not in dp:
                if houses[index] == 0:
                    # House not colored
                    dp[key] = min(dfs(index+1, target - (new_color != prev_color), new_color) + cost[index][new_color-1] for new_color in range(1, n+1))
                else:
                    # House already colored
                    dp[key] = dfs(index+1, target - (houses[index] != prev_color), houses[index])
            return dp[key]
        result = dfs(0, target, -1)
        return result if result < float("inf") else -1