256. Paint House

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.

class Solution {

public:

    /*

    int minCost(vector<vector<int>>& costs) {

        if (costs.size() == 0) {

            return 0;

        }

        int size = costs[0].size();

        vector<int> helper = costs[0];

        for (int i = 1; i < costs.size(); i++) {

            vector<int> tmp(size, 0);

            tmp[0] = costs[i][0] + min(helper[1], helper[2]);

            tmp[1] = costs[i][1] + min(helper[0], helper[2]);

            tmp[2] = costs[i][2] + min(helper[1], helper[0]);

            helper = tmp;

            tmp.clear();

        }

        

        int ans = INT_MAX;

        for (auto num : helper) {

            ans = min(ans, num);

        }

        return ans;

    }

    */

    int minCost(vector<vector<int>>& costs) {

        if (costs.size() == 0) {

            return 0;

        }

        int size = costs[0].size();

        for (int i = 1; i < costs.size(); i++) {

            vector<int> tmp(size, 0);

            costs[i][0] = costs[i][0] + min(costs[i-1][1], costs[i-1][2]);

            costs[i][1] = costs[i][1] + min(costs[i-1][0], costs[i-1][2]);

            costs[i][2] = costs[i][2] + min(costs[i-1][1], costs[i-1][0]);

        }

        

        int ans = INT_MAX;

        for (auto num : costs[costs.size() - 1]) {

            ans = min(ans, num);

        }

        return ans;

    }

};

class Solution {
public:
 int minCost(vector<vector<int>>& costs) {
  if(costs.empty()) return 0;
  int red=costs[0][0];
  int blue=costs[0][1];
  int green=costs[0][2];
  for(int i=1;i<costs.size();i++){
   int t_red=red;
   int t_blue=blue;
   int t_green=green;
   red=min(t_blue,t_green)+costs[i][0];
   blue=min(t_red,t_green)+costs[i][1];
   green=min(t_red,t_blue)+costs[i][2];

  }
  return min(green,min(red,blue));   
 }
};