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;
}
}