Paint house

/*
Logic:
dp array rows represents houses and 3 columns for red,blue and green colors.
dp[i][j] will store the minimum cost by coloring ith house by jth color ie. j=0->red,j=1->blue,j=2->green
coloring of first house with jth colors will be costs[0][j]
so,if we color ith house with red,then minimum cost till ith house will be min of cost of previous house
colored with blue and green because same color can't be in adjacent house and same goes for other colors also.
finally,in last row,we will have minimum cost coloring last house with red,blue and green,return their minimum.
*/

int Solution::solve(vector<vector<int> > &costs) {
        int n=costs.size();
        vector<vector<int> > dp(n, vector<int>(3, 0));
        dp[0][0]=costs[0][0];
        dp[0][1]=costs[0][1];
        dp[0][2]=costs[0][2];
        for(int i=1;i<costs.size();i++){
            dp[i][0]=min(dp[i-1][1],dp[i-1][2])+costs[i][0];
            dp[i][1]=min(dp[i-1][0],dp[i-1][2])+costs[i][1];
            dp[i][2]=min(dp[i-1][0],dp[i-1][1])+costs[i][2];
        }
        n--;
        return min(min(dp[n][0],dp[n][1]),dp[n][2]);
}