1423. Maximum Points You Can Obtain from Cards ------M

 There are several cards arranged in a row, and each card has an associated number of points The points are given in the integer array cardPoints.

In one step, you can take one card from the beginning or from the end of the row. You have to take exactly k cards.

Your score is the sum of the points of the cards you have taken.

Given the integer array cardPoints and the integer k, return the maximum score you can obtain.

 

Example 1:

Input: cardPoints = [1,2,3,4,5,6,1], k = 3
Output: 12
Explanation: After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.

Example 2:

Input: cardPoints = [2,2,2], k = 2
Output: 4
Explanation: Regardless of which two cards you take, your score will always be 4.

Example 3:

Input: cardPoints = [9,7,7,9,7,7,9], k = 7
Output: 55
Explanation: You have to take all the cards. Your score is the sum of points of all cards.

Example 4:

Input: cardPoints = [1,1000,1], k = 1
Output: 1
Explanation: You cannot take the card in the middle. Your best score is 1. 

Example 5:

Input: cardPoints = [1,79,80,1,1,1,200,1], k = 3
Output: 202

 

Constraints:

• 1 <= cardPoints.length <= 10^5
• 1 <= cardPoints[i] <= 10^4
• 1 <= k <= cardPoints.length

A:

首先试验  backtrack with memoization

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class Solution { public: int maxScore(vector<int>& cardPoints, int k) { unordered_map<long, int> M; return helper(cardPoints, 0, cardPoints.size()-1, k, M); } private: int helper(vector<int>& arr, int start, int end, int k, unordered_map<long, int>& M){ long key = start*10001+end; if(M.find(key) != M.end()){ return M[key]; } if(k==0){ return 0; } int res1 = arr[start] + helper(arr, start+1, end, k-1,M); int res2 = arr[end] + helper(arr, start, end-1, k-1,M); M[key] = max(res1,res2); return M[key]; } };

结果竟然超时了。

进一步的修改，是把cardPoints两边各k个取下来。

然后找max_subarray_sum, with length k

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class Solution { public: int maxScore(vector<int>& cardPoints, int k) { int n = cardPoints.size(); k = min(k,n); // detach the vector, making them as the min vector<int> V{0}; // attach first k for(int i =0;i<k;i++){ V.push_back(cardPoints[k-1-i]); } //attach backward of last k for(int d =1;d<=k;d++){ V.push_back(cardPoints[n-d]); } int maxRes = INT_MIN; for(int i =1;i<V.size();i++){ V[i] += V[i-1]; if(i>=k){ maxRes = max(maxRes, V[i] - V[i-k]); } } return maxRes; } };

这里没想到的一点是： 加前面k的数字的时候，没有倒着加入 （ see line 10)

然而只beat了22%的

进一步优化。我们其实并不需要真正地开辟数组V.

而只需要计算其subarry 的 sum 即可。

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class Solution { public: int maxScore(vector<int>& cardPoints, int k) { int n = cardPoints.size(); k = min(n,k); int total = 0; for(int i =0;i< k;i++){ total += cardPoints[i]; } int res = total; for(int d = 1; d<=k; d++){ total += cardPoints[n-d] - cardPoints[k-d]; res = max(total, res); } return res; } };