Maximum Points You Can Obtain from Cards

Intuition

Whatever combination of front and back picks you make, the leftover cards are always a single contiguous run in the middle whose length is exactly n minus k. The score you keep equals the total of all cards minus that leftover run. So rather than juggling two ends, slide a fixed-width window across the array to find the run with the smallest sum, and the maximum score is simply the total minus that minimum.

Algorithm walkthrough

1. 1Compute the sum of every card, and let the leftover window width be n minus k.
2. 2If that width is zero you must take every card, so return the total directly.
3. 3Sum the first window of that width to seed the current and minimum leftover sums.
4. 4Slide the window one step at a time, adding the entering card and removing the leaving card, tracking the smallest leftover sum.
5. 5Return the total minus the smallest leftover sum found.

Implementation

class Solution:
    def maxScore(self, cardPoints: List[int], k: int) -> int:
        n = len(cardPoints)
        window = n - k
        total = sum(cardPoints)
        if window == 0:
            return total
        current = sum(cardPoints[:window])
        min_window = current
        for i in range(window, n):
            current += cardPoints[i] - cardPoints[i - window]
            min_window = min(min_window, current)
        return total - min_window