265. Paint House II 房屋涂不同颜色的油漆

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 an n x k cost matrix costs.

• 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...

Return the minimum cost to paint all houses.

 

Example 1:

Input: costs = [[1,5,3],[2,9,4]]
Output: 5
Explanation:
Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5; 
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.

Example 2:

Input: costs = [[1,3],[2,4]]
Output: 5

参考：https://leetcode.com/problems/paint-house-ii/discuss/69502/Evolve-from-brute-force-to-optimal

This is similar to paint house.

 

1. O((k-1)^n) brute force

    int minCostII(vector<vector<int>>& costs) {
        if(costs.empty()) return 0;
        return minCost(-1,-1,costs);        
    }
    int minCost(int i,int j, vector<vector<int>>& costs) { //minCost starting from house i with color j
        if(i==costs.size()) return 0;
        int mc = INT_MAX;
        for(int k=0;k<costs[0].size();k++) if(k!=j) mc = min(mc, minCost(i+1,k,costs));
        return i<0? mc : mc+costs[i][j];
    }

2. O(nk^2) Memoization

    int minCostII(vector<vector<int>>& costs) {
        if(costs.empty()) return 0;
        vector<vector<int>> mem(costs.size(),vector<int>(costs[0].size()));
        return minCost(-1,-1,mem,costs);        
    }
    int minCost(int i,int j, vector<vector<int>>& mem, vector<vector<int>>& costs) {
        if(i==costs.size()) return 0;
        if(i>0 && mem[i][j]) return mem[i][j];
        int mc = INT_MAX;
        for(int k=0;k<costs[0].size();k++) if(k!=j) mc = min(mc, minCost(i+1,k,mem,costs));
        return i<0? mc : mem[i][j]=mc+costs[i][j];
    }

3. O(nk^2) dp

    int minCostII(vector<vector<int>>& costs) {
        if(costs.empty()) return 0;
        int n = costs.size(), k = costs[0].size();
        vector<vector<int>> dp(n+1,vector<int>(k));
        for(int i=n-1;i>=0;i--)
            for(int j=0;j<k;j++)
                dp[i][j]=getMin(j,dp[i+1]) + costs[i][j];
        return getMin(-1, dp[0]);        
    }
    int getMin(int j, vector<int> &pre) {
        int mc = INT_MAX;
        for(int i=0;i<pre.size();i++) if(i!=j) mc = min(mc,pre[i]);
        return mc;
    }

4. O(nk) dp

    int minCostII(vector<vector<int>>& costs) {
        int pre1=0,pre2=0,c1=-1;
        for(auto &v:costs) {
            int cur1=INT_MAX,cur2,co1;
            for(int i=0;i<v.size();i++) {
                int c = v[i]+ (i==c1?pre2:pre1);
                if(c<cur1) {
                    cur2 = cur1;
                    co1 = i;
                    cur1 = c;
                } else if (c<cur2) cur2 = c;
            }
            pre1 = cur1;
            pre2 = cur2;
            c1 = co1;
        }
        return pre1;
    }