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Maximum Points You Can Obtain from Cards

Medium·LeetCode #1423 Pro OnlyInterview-Level
LeetCode
01

Problem Statement

Some cards are laid out in a row, and card i is worth cardPoints[i] points.

You make exactly k moves. In each move you take one card from either end of the row — the leftmost or the rightmost card that is still there. Your score is the total of the cards you took.

Return the highest score you can get.

Input

An integer array cardPoints and an integer k.

Output

The maximum total points from taking exactly k cards, each from the left or right end.

Example 1
Input: cardPoints = [1, 2, 3, 4, 5, 6, 1], k = 3
Output: 12

Take the three rightmost cards: 1 + 6 + 5 = 12. No other choice does better.

Example 2
Input: cardPoints = [2, 2, 2], k = 2
Output: 4

Any two cards are worth 2 + 2 = 4.

Example 3
Input: cardPoints = [9, 7, 7, 9, 7, 7, 9], k = 7
Output: 55

k equals the number of cards, so you take all of them: the total is 55.

Constraints
  • • 1 ≤ cardPoints.length ≤ 10^5
  • • 1 ≤ cardPoints[i] ≤ 10^4
  • • 1 ≤ k ≤ cardPoints.length
02

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“take from the beginning or the end”
→ Order doesn't matter; you end with a prefix + a suffix. What remains is a contiguous middle block.
“exactly k cards”
→ The leftover block always has exactly n − k cards → fixed window size.
“maximum score”
→ Max taken = total − min left → minimise the window sum.
03

Approach Complexity Ladder

Brute Force — try every left/right splitO(K²)O(1)

For each i in 0..k, add the first i and last k − i cards from scratch

Optimal — minimise the leftover middle windowO(N)O(1)

Total minus the smallest sum of a size n − k window

Brute Force Intuition

Any sequence of moves ends with some i cards taken from the left and k − i from the right. Try every i from 0 to k, add up those cards from scratch, and keep the best total.

Recognition rule for this sub-pattern:

Window size is fixed at K. Each slide adds one element on the right and removes one on the left — update the aggregate in O(1).
Remember this · Sliding Window
Trigger
Contiguous subarray / substring + longest, shortest, maximum, count, or "of size K".
Think
Can I keep a valid range and move it, instead of checking every subarray from scratch?
Key question
When is my current window valid?
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