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.
Example:
Input: [[17,2,17],[16,16,5],[14,3,19]] Output: 10 Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. Minimum cost: 2 + 5 + 3 = 10.
// 4ms, 97%
class Solution {
public:
int minCost(vector<vector<int>>& costs) {
vector<vector<int>> dp(costs.size()+1, vector<int>(3, 0));
const auto N = costs.size();
for (int i = 1; i <= N; i++) {
dp[i][0] = costs[i-1][0] + min(dp[i-1][1], dp[i-1][2]);
dp[i][1] = costs[i-1][1] + min(dp[i-1][0], dp[i-1][2]);
dp[i][2] = costs[i-1][2] + min(dp[i-1][0], dp[i-1][1]);
}
return min(dp[N][0], min(dp[N][1], dp[N][2]));
}
};