[https://leetcode.com/problems/paint-house-ii/](https://leetcode.com/problems/paint-house-ii/)↵
↵
There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.↵
↵
I am unable to pass all the test cases↵
↵
```↵
class Solution {↵
public:↵
int minCostII(vector<vector<int>>& cost) {↵
if(cost.size()==0)return 0;↵
int n=cost.size();↵
int k=cost[0].size();↵
↵
int dp[n][k];↵
int min1=INT_MAX;↵
int min2=INT_MAX;↵
int minind=-1;↵
for(int i=0;i<k;i++){↵
dp[0][i]=cost[0][i];↵
if(cost[0][i]<min1){↵
min2=min1;↵
minind=i;↵
min1=cost[0][i];↵
}else min2=min(min2,cost[0][i]);↵
↵
↵
}↵
cout<<min2;↵
for(int i=1;i<n;i++){↵
int curmin=INT_MAX;↵
int curmin2=INT_MAX;↵
int curind=0;↵
for(int j=0;j<k;j++){↵
if(minind!=j){↵
dp[i][j]=min1+cost[i][j];↵
}else{↵
dp[i][j]=min2+cost[i][j];↵
}↵
if(curmin>dp[i][j]){↵
curmin2=curmin;↵
curmin=dp[i][j];↵
minind=j;↵
↵
}else curmin2=min(curmin2,dp[i][j]);↵
↵
}↵
min1=curmin;↵
min2=curmin2;↵
minind=curind;↵
}↵
↵
return min1;↵
}↵
};↵
```
↵
There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.↵
↵
I am unable to pass all the test cases↵
↵
```↵
class Solution {↵
public:↵
int minCostII(vector<vector<int>>& cost) {↵
if(cost.size()==0)return 0;↵
int n=cost.size();↵
int k=cost[0].size();↵
↵
int dp[n][k];↵
int min1=INT_MAX;↵
int min2=INT_MAX;↵
int minind=-1;↵
for(int i=0;i<k;i++){↵
dp[0][i]=cost[0][i];↵
if(cost[0][i]<min1){↵
min2=min1;↵
minind=i;↵
min1=cost[0][i];↵
}else min2=min(min2,cost[0][i]);↵
↵
↵
}↵
cout<<min2;↵
for(int i=1;i<n;i++){↵
int curmin=INT_MAX;↵
int curmin2=INT_MAX;↵
int curind=0;↵
for(int j=0;j<k;j++){↵
if(minind!=j){↵
dp[i][j]=min1+cost[i][j];↵
}else{↵
dp[i][j]=min2+cost[i][j];↵
}↵
if(curmin>dp[i][j]){↵
curmin2=curmin;↵
curmin=dp[i][j];↵
minind=j;↵
↵
}else curmin2=min(curmin2,dp[i][j]);↵
↵
}↵
min1=curmin;↵
min2=curmin2;↵
minind=curind;↵
}↵
↵
return min1;↵
}↵
};↵
```
