Maximum Points You Can Obtain from Cards — LeetCode

Description

Several cards are 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.

Constraints:

• 1 <= cardPoints.length <= 105
• 1 <= cardPoints[i] <= 104
• 1 <= k <= cardPoints.length

Solution

We use a sliding window approach to find the best window having the maximum score

We can start by iterating through the first K cards of the array and finding the total points. At this point, our reverse window will be the cards from i = K to j = C.length — 1. At each iteration, we’ll slide the window backwards, removing one card from the left side (-C[i]) and adding one card from the right side (+C[j]) each time.

We should keep track of the best possible result at each iteration, and then return the best once we reach the end.

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

• Time Complexity: O(k)
• Space Complexity: O(1)