Paint House I

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.

O(nk) time O(n) space

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length < 1) return 0;
        int length = costs.length + 1;
        int[][] total = new int[length][costs[0].length];
        for (int i = 0; i < 3; i++) {
            total[0][i] = 0;
        }
        
        for (int j = 1; j < length; j++) {
            for (int c = 0; c < 3; c++) {
                total[j][c] = costs[j - 1][c] + getPrevMin(total, j - 1, c);
            }
        }
        return Math.min(total[length - 1][2], Math.min(total[length - 1][0], total[length - 1][1]));
    }
    
    private int getPrevMin(int[][] total, int house, int color) {
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < total[0].length; i++) {
            if (i == color && i == 0 && total[0].length - 1 == color) {
                return 0;
            } else if (i == color) {
                continue;
            } else {
                min = Math.min(min, total[house][i]);
            }
        }
        return min;
    }
}

O(nk) time and O(1) space

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
// starting from the second house/row, costs[i][0] represents the cost of painting this house red plus the previous house, etc
public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length < 1){
            return 0;
        }
        int length = costs.length, r = 0, b = 0, g = 0;
        
        for (int i = 0; i < length; i++){
            int rr = r, bb = b, gg = g;
            r = costs[i][0] + Math.min(bb, gg);
            g = costs[i][1] + Math.min(rr, bb);
            b = costs[i][2] + Math.min(rr, gg);
        }
        
        return Math.min(r, Math.min(g, b));
    }
}