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# Solution: Maximum Points You Can Obtain from Cards

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

#### Description:

(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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.

#### Examples:

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`

#### Idea:

(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

Since we're forced to take K amount of cards no matter what, we can solve this problem with a two-pointer system with a sliding window approach. Instead of counting the sum of the values between the two pointers, we'll instead be counting the sum of the values outside the sliding window.

We can start by iterating through the first K cards of our card list (C) 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, then return best once we reach the end.

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

#### Javascript Code:

``````var maxScore = function(C, K) {
let total = 0
for (let i = 0; i < K; i++) total += C[i]
let best = total
for (let i = K - 1, j = C.length - 1; ~i; i--, j--)
total += C[j] - C[i], best = Math.max(best, total)
return best
};
``````

#### Python Code:

``````class Solution:
def maxScore(self, C: List[int], K: int) -> int:
best = total = sum(C[:K])
for i in range (K-1, -1, -1):
total += C[i + len(C) - K] - C[i]
best = max(best, total)
return best
``````

#### Java Code:

``````class Solution {
public int maxScore(int[] C, int K) {
int total = 0;
for (int i = 0; i < K; i++) total += C[i];
int best = total;
for (int i = K - 1, j = C.length - 1; i >= 0; i--, j--) {
total += C[j] - C[i];
best = Math.max(best, total);
}
return best;
}
}
``````

#### C++ Code:

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

## Discussion (1)

Rohith V

Sharing my Java Solution

``````class Solution {
public int maxScore(int[] cardPoints, int k) {
int length = cardPoints.length;
if (k == length) {
int sum = 0;
for (int number : cardPoints) {
sum += number;
}
return sum;
}
int leftSum = 0;
int rightSum = 0;
for (int i=0; i<k; i++) {
leftSum += cardPoints[i];
}
int maxSum = leftSum;
for (int i=0; i<k; i++) {
leftSum -= cardPoints[k - i - 1];
rightSum += cardPoints[length - i - 1];
maxSum = Math.max(maxSum, leftSum + rightSum);
}
return maxSum;
}
}
``````