LeetCode "Paint House II"

Classic DP! For house[i] to pick a min-color from dp[i-1], we only need to check if color j is the min cost color index of dp[i - 1]; if yes, then we pick the 2nd smallest min cost color. Also: we can optimize memory complexity by using rolling array.

class Solution 
{    
public:
    void getSm1n2(vector<int> &house, int &sm1, int &sm2)
    {
        size_t len = house.size();

        int curr1 = house[0], curr2 = INT_MAX;
        sm1 = 0;

        for (int i = 1; i < len; i++)
        {
            int v = house[i];
            if (v < curr1)
            {
                curr2 = curr1;
                sm2 = sm1;
                curr1 = v;
                sm1 = i;
            }
            else
            {
                if (v < curr2)
                {
                    curr2 = v;
                    sm2 = i;
                }
            }
        }
    }

    int minCostII(vector<vector<int>>& costs) 
    {
        int ret = 0;
        unsigned nHouseCnt = costs.size();
        if (nHouseCnt == 0) return ret;
        unsigned nColorCnt = costs[0].size();

        //    Can be optimized using rolling array
        vector<vector<int>> dp(nHouseCnt, vector<int>(nColorCnt, 0));
        dp[0] = costs[0];
        int sm1, sm2;
        getSm1n2(dp[0], sm1, sm2);
        for (int i = 1; i < nHouseCnt; i++)
        {
            for (int j = 0; j < nColorCnt; j++)
            {
                dp[i][j] = dp[i - 1][j == sm1 ? sm2 : sm1] + costs[i][j];
            }
            getSm1n2(dp[i], sm1, sm2);
        }
        ret = *std::min_element(dp[nHouseCnt - 1].begin(), dp[nHouseCnt - 1].end());
        return ret;
    }
};

And I got 1AC in my onsite interview w this problem :)