Painting with minimum cost

    int minCostPaintAtMostTwoSameColor(int[][] cost)
    {
        // cost in form n x 3
        // cost[0][1] = paint house 0 with color 1
        int n = cost.length, inf = Integer.MAX_VALUE;
        // f[i][j] = minimum cost to paint up to i houses ending with paint j
        int[][] f = new int[n + 1][3];
        f[1] = cost[0];
        for(int j = 0; j < 3; j++)
            f[2][j] = Math.min(f[1][0], Math.min(f[1][1], f[1][2])) + cost[1][j];
        for(int i = 2; i < n; i++)
        {
            // loop through colors
            for(int j = 0; j < 3; j++)
            {
                int diffColor = inf, sameColor = inf;
                for(int k = 0; j < 3; k++)
                {
                    if(j == k) continue;
                    diffColor = Math.min(diffColor, f[i][k]);
                    sameColor = Math.min(sameColor, f[i - 1][k] + cost[i][j]);
                }
                f[i + 1][j] = Math.min(diffColor, sameColor);
            }
        }
        return Math.min(Math.min(f[n][0], f[n][1]), f[n][2]);
    }

    int minCostPaintNoSameColor(int[][] cost)
    {
        int n = cost.length, inf = Integer.MAX_VALUE;
        int[][] f = new int[n + 1][3];
        for(int i = 0; i < n; i++)
        {
            f[i + 1][0] = Math.min(f[i][1], f[i][2]) + cost[i][0];
            f[i + 1][1] = Math.min(f[i][0], f[i][2]) + cost[i][1];
            f[i + 1][2] = Math.min(f[i][0], f[i][1]) + cost[i][2];
        }
        return Math.min(Math.min(f[n][0], f[n][1]), f[n][2]);
    }