Paint House

Paint House dynamic programming

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.

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.

Note: All costs are positive integers.

Paint House Solution

class PaintHouse {
    public int minCost(int[][] costs) {
        if(costs == null || costs.length == 0) {
            return 0;
        }

        for(int i = 1; i < costs.length; 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][0], costs[i - 1][1]);
        }

        return Math.min(Math.min(costs[costs.length - 1][0], costs[costs.length - 1][1]), costs[costs.length - 1][2]);
    }
}
Last modified October 4, 2020