Description

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

DP, time O(n), space O(n)

class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0 || costs[0].length == 0) {
            return 0;
        }
        
        int n = costs.length;
        int[][] costSum = new int[n + 1][3];
        
        for (int i = 0; i < n; ++i) {
            costSum[i + 1][0] = costs[i][0] 
                + Math.min(costSum[i][1], costSum[i][2]);
            costSum[i + 1][1] = costs[i][1] 
                + Math.min(costSum[i][0], costSum[i][2]);
            costSum[i + 1][2] = costs[i][2] 
                + Math.min(costSum[i][0], costSum[i][1]);
        }
        
        return min(costSum[n]);
    }
    
    private int min(int[] nums) {
        int min = Integer.MAX_VALUE;
        
        for (int n : nums) {
            min = Math.min(n, min);
        }
        
        return min;
    }
}