1473 - Paint House III.

Hard

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

• <code>1 <= costj<= 10⁴</code>