256. Paint House

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.

Solution:

We use mincost[i][j] to denote the minimum cost at house i using paint j.

Therefore, since we cannot use the same paint in two adjacent houses,

mincost[i][j] = costs[i][j] + min(mincost[i][k], k ≠ j)

We can optimized it by using only one cost array.

costs[i][j] = costs[i][j] + min(costs[i][k], k ≠ j)



Code:

public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) {
            return 0;
        }
        if (costs[0] == null || costs[0].length == 0) {
            return 0;
        }
        for (int i = 1; i < costs.length; i++) {
            costs[i][0] = costs[i][0] + Math.min(costs[i - 1][1], costs[i - 1][2]);
            costs[i][1] = costs[i][1] + Math.min(costs[i - 1][0], costs[i - 1][2]);
            costs[i][2] = costs[i][2] + Math.min(costs[i - 1][0], costs[i - 1][1]);
        }
        int n = costs.length;
        return Math.min(costs[n - 1][0], Math.min(costs[n - 1][1], costs[n - 1][2]));
    }
}