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Leetcode Problem #1423 (Medium): Maximum Points You Can Obtain from Cards

Description:

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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:

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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:

(Jump to: Problem Description || Solution Idea)

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:

(Jump to: Problem Description || Solution Idea)

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:

(Jump to: Problem Description || Solution Idea)

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:

(Jump to: Problem Description || Solution Idea)

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;
    }
};