Problem 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.
An interesting DP problem. This link posts a nice solution which gives costs[i][j] a new meaning and modify it directly and so save the usage of additional spaces.
Well, personally I would like to keep costs unmodified. I rewrite the code in C++, a little verbose than the one in the above link:-)
1 class Solution { 2 public: 3 int minCost(vector<vector<int>>& costs) { 4 if (costs.empty()) return 0; 5 int n = costs.size(), r = 0, g = 0, b = 0; 6 for (int i = 0; i < n; i++) { 7 int rr = r, bb = b, gg = g; 8 r = costs[i][0] + min(bb, gg); 9 b = costs[i][1] + min(rr, gg); 10 g = costs[i][2] + min(rr, bb); 11 } 12 return min(r, min(b, g)); 13 } 14 };
r/b/g in the i-th loop means the minimum costs to paint the i-th house in red/blue/green respectively plus painting the previous houses. The time and space complexities are still ofO(n) and O(1).