Paint House III

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 has 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 * n matrix cost and an integer target where:

• houses[i]: is the color of the house i, 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 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 = [0,0,0,0,0], cost = [[1,10],[10,1],[1,10],[10,1],[1,10]], m = 5, n = 2, target = 5
Output: 5

Example 4:

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] <= 10^4

class Solution {
    Integer[][][] dp;

    public int minCost(int[] houses, int[][] cost, int m, int n, int target) {
        dp = new Integer[m][n][target + 1];
        return helper(houses, cost, 0, -1, target);
    }

    public int helper(int[] house, int[][] cost, int index, int colorOfPrevHouse, int target) {
        if (index == house.length) {
            if (target == 0)
                return 0;
            return -1;
        }
        if (target < 0)
            return -1;
        if (colorOfPrevHouse != -1)
            if (dp[index][colorOfPrevHouse][target] != null)
                return dp[index][colorOfPrevHouse][target];
        int ans = Integer.MAX_VALUE;
        // If current house is already colored
        if (house[index] != 0) {
            // If current house has same color as previous house
            if (house[index] == colorOfPrevHouse + 1)
                ans = helper(house, cost, index + 1, colorOfPrevHouse, target);
            else if (target > 0)
                ans = helper(house, cost, index + 1, house[index] - 1, target - 1);
        }
        // Else
        else {
            if (colorOfPrevHouse != -1) {
                // If we paint same color as previous house
                int ans2 = helper(house, cost, index + 1, colorOfPrevHouse, target);
                if (ans2 != -1)
                    ans = Math.min(ans, ans2 + cost[index][colorOfPrevHouse]);
                // Else if we want it in a different color
                if (target > 0) {
                    for (int j = 0; j < cost[index].length; j++) {
                        if (j == colorOfPrevHouse)
                            continue;
                        int res = helper(house, cost, index + 1, j, target - 1);
                        if (res != -1)
                            ans = Math.min(ans, cost[index][j] + res);
                    }
                }
            } else
                for (int j = 0; j < cost[index].length; j++) {
                    int res = helper(house, cost, index + 1, j, target - 1);
                    if (res != -1)
                        ans = Math.min(cost[index][j] + res, ans);
                }
        }
        if (colorOfPrevHouse != -1)
            dp[index][colorOfPrevHouse][target] = ans == Integer.MAX_VALUE ? -1 : ans;
        return ans == Integer.MAX_VALUE ? -1 : ans;
    }
}