package com.freetymekiyan.algorithms.level.medium;
/**
* There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. The cost of
* painting each house with a certain color is different. You have to paint all the houses such that no two adjacent
* houses have the same color.
* <p>
* The cost of painting each house with a certain color is represented by a n x 3 cost matrix. For example, costs[0][0]
* is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so
* on... Find the minimum cost to paint all houses.
* <p>
* Note:
* All costs are positive integers.
* <p>
* Company Tags: LinkedIn
* Tags: Dynamic Programming
* Similar Problems: (E) House Robber, (M) House Robber II, (H) Paint House II, (E) Paint Fence
*/
public class PaintHouse {
/**
* DP.
* Recurrence Relation:
* The color of current house i has 3 possibilities.
* If it's red, the min cost will be cost of red at i + min(cost of blue at i-1, cost of green at i-1).
* That is: costs[i][0] + min(costs[i - 1][1], costs[i - 1][2]).
* The same goes for if it's blue or green.
* So generate all possibilities.
* Base case:
* If costs matrix is empty, return 0.
*/
public int minCost(int[][] costs) {
if (costs.length == 0) {
return 0;
}
int n = costs.length;
for (int i = 1; i < n; i++) {
costs[i][0] += Math.min(costs[i - 1][1], costs[i - 1][2]);
costs[i][1] += Math.min(costs[i - 1][0], costs[i - 1][2]);
costs[i][2] += Math.min(costs[i - 1][1], costs[i - 1][0]);
}
return Math.min(Math.min(costs[n - 1][0], costs[n - 1][1]), costs[n - 1][2]);
}
}