- Difficulty: Hard
- Tags: LeetCode, Hard, Array, Dynamic Programming, leetcode-1473, O(m * t * n^2), O(t * n)
Problem
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]
contains5
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 housei
, and0
if the house is not painted yet.cost[i][j]
: is the cost of paint the housei
with the colorj + 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