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1423. Maximum Points You Can Obtain from Cards

Description

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.

 

Constraints:

  • 1 <= cardPoints.length <= 105
  • 1 <= cardPoints[i] <= 104
  • 1 <= k <= cardPoints.length

Solutions

Solution 1: Sliding Window

We can use a sliding window of length $k$ to simulate this process.

Initially, we place the window at the end of the array, i.e., the $k$ positions from index $n-k$ to index $n-1$. The sum of the points of the cards in the window is denoted as $s$, and the initial value of the answer $ans$ is also $s$.

Next, we consider the situation where we take $1, 2, …, k$ cards from the beginning of the array in turn. Suppose the card taken is $cardPoints[i]$. Then we add it to $s$. Due to the length limit of the window is $k$, we need to subtract $cardPoints[n-k+i]$ from $s$. In this way, we can calculate the sum of the points of the $k$ cards taken and update the answer $ans$.

The time complexity is $O(k)$, where $k$ is the integer given in the problem. The space complexity is $O(1)$.

  • class Solution {
    public int maxScore(int[] cardPoints, int k) {
        int s = 0, n = cardPoints.length;
        for (int i = n - k; i < n; ++i) {
            s += cardPoints[i];
        }
        int ans = s;
        for (int i = 0; i < k; ++i) {
            s += cardPoints[i] - cardPoints[n - k + i];
            ans = Math.max(ans, s);
        }
        return ans;
    }
}

  • class Solution {
public:
    int maxScore(vector<int>& cardPoints, int k) {
        int n = cardPoints.size();
        int s = accumulate(cardPoints.end() - k, cardPoints.end(), 0);
        int ans = s;
        for (int i = 0; i < k; ++i) {
            s += cardPoints[i] - cardPoints[n - k + i];
            ans = max(ans, s);
        }
        return ans;
    }
};

  • class Solution:
    def maxScore(self, cardPoints: List[int], k: int) -> int:
        ans = s = sum(cardPoints[-k:])
        for i, x in enumerate(cardPoints[:k]):
            s += x - cardPoints[-k + i]
            ans = max(ans, s)
        return ans


  • func maxScore(cardPoints []int, k int) int {
	n := len(cardPoints)
	s := 0
	for _, x := range cardPoints[n-k:] {
		s += x
	}
	ans := s
	for i := 0; i < k; i++ {
		s += cardPoints[i] - cardPoints[n-k+i]
		ans = max(ans, s)
	}
	return ans
}

  • function maxScore(cardPoints: number[], k: number): number {
    const n = cardPoints.length;
    let s = cardPoints.slice(-k).reduce((a, b) => a + b);
    let ans = s;
    for (let i = 0; i < k; ++i) {
        s += cardPoints[i] - cardPoints[n - k + i];
        ans = Math.max(ans, s);
    }
    return ans;
}


  • /**
 * @param {number[]} cardPoints
 * @param {number} k
 * @return {number}
 */
var maxScore = function (cardPoints, k) {
    const n = cardPoints.length;
    let s = cardPoints.slice(-k).reduce((a, b) => a + b);
    let ans = s;
    for (let i = 0; i < k; ++i) {
        s += cardPoints[i] - cardPoints[n - k + i];
        ans = Math.max(ans, s);
    }
    return ans;
};


  • class Solution {
    /**
     * @param Integer[] $cardPoints
     * @param Integer $k
     * @return Integer
     */
    function maxScore($cardPoints, $k) {
        $n = count($cardPoints);
        $s = array_sum(array_slice($cardPoints, -$k));
        $ans = $s;
        for ($i = 0; $i < $k; ++$i) {
            $s += $cardPoints[$i] - $cardPoints[$n - $k + $i];
            $ans = max($ans, $s);
        }
        return $ans;
    }
}


  • # @param {Integer[]} card_points
# @param {Integer} k
# @return {Integer}
def max_score(card_points, k)
  n = card_points.length
  s = card_points[-k..].sum
  ans = s
  k.times do |i|
    s += card_points[i] - card_points[n - k + i]
    ans = [ans, s].max
  end
  ans
end


  • object Solution {
    def maxScore(cardPoints: Array[Int], k: Int): Int = {
        val n = cardPoints.length
        var s = cardPoints.takeRight(k).sum
        var ans = s
        for (i <- 0 until k) {
            s += cardPoints(i) - cardPoints(n - k + i)
            ans = ans.max(s)
        }
        ans
    }
}


  • public class Solution {
    public int MaxScore(int[] cardPoints, int k) {
        int n = cardPoints.Length;
        int s = cardPoints[^k..].Sum();
        int ans = s;
        for (int i = 0; i < k; ++i) {
            s += cardPoints[i] - cardPoints[n - k + i];
            ans = Math.Max(ans, s);
        }
        return ans;
    }
}


  • impl Solution {
    pub fn max_score(card_points: Vec<i32>, k: i32) -> i32 {
        let n = card_points.len();
        let k = k as usize;
        let mut s: i32 = card_points[n - k..].iter().sum();
        let mut ans: i32 = s;
        for i in 0..k {
            s += card_points[i] - card_points[n - k + i];
            ans = ans.max(s);
        }
        ans
    }
}


  • class Solution {
    func maxScore(_ cardPoints: [Int], _ k: Int) -> Int {
        let n = cardPoints.count
        var s = cardPoints.suffix(k).reduce(0, +)
        var ans = s
        for i in 0..<k {
            s += cardPoints[i] - cardPoints[n - k + i]
            ans = max(ans, s)
        }
        return ans
    }
}

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