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Kotlin-based LeetCode algorithm problem solutions, regularly updated

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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`
*   1 <= cost[i][j] <= 10⁴