1423. Maximum Points You Can Obtain from Cards

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Problem

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

Discussion

The problem requires to find the maximum sum of k integers, which are being picked from the beginning or the end only.

Solving this problem, we can list out the combinations of the k integers and finding the maximum in them. Or, we can find the minimum of the unpicked integers by sliding windows, and give the difference from sum of all integers.

Solution

The solution begins with finding the sum of all integers in an O(n) loop. Within the loop, we manage the sliding windows by summing the first n - k integers as window.

After the first n - k integers, we update the sliding window by dropping the first integer (card[i - len(card) + k]) and adding a new integer after the window (card[i]). And we update the minimum of the sliding window at each time.

from typing import List

class Solution:
    def maxScore(self, cardPoints: List[int], k: int) -> int:
        sum, window, min_window = 0, 0, 0

        for i in range(len(cardPoints)):
            sum += cardPoints[i]
            if i < len(cardPoints) - k:
                window += cardPoints[i]
                min_window = window
            else:
                window = window + \
                    cardPoints[i] - cardPoints[i - len(cardPoints) + k]
                min_window = min(min_window, window)

        return sum - min_window

Complexity Analysis

Time Complexity: O(n), as the solution is completed within an O(n) loop.
Space Complexity: O(1), as we only cache three variables and updating them in each iteration.

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1423. Maximum Points You Can Obtain from Cards

Problem

Discussion

Solution

Complexity Analysis