Skip to content

Latest commit

 

History

History
332 lines (275 loc) · 8.07 KB

README_EN.md

File metadata and controls

332 lines (275 loc) · 8.07 KB
comments difficulty edit_url rating source tags
true
Medium
1573
Weekly Contest 186 Q2
Array
Prefix Sum
Sliding Window

1423. Maximum Points You Can Obtain from Cards

中文文档

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 $O(1)$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

JavaScript

/**
 * @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;
};

C#

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;
    }
}

PHP

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;
    }
}

Scala

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
    }
}

Swift

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
    }
}

Ruby

# @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

Kotlin

class Solution {
    fun maxScore(cardPoints: IntArray, k: Int): Int {
        val n = cardPoints.size
        var s = cardPoints.sliceArray(n - k until n).sum()
        var ans = s
        for (i in 0 until k) {
            s += cardPoints[i] - cardPoints[n - k + i]
            ans = maxOf(ans, s)
        }
        return ans
    }
}

Dart

class Solution {
  int maxScore(List<int> cardPoints, int k) {
    int n = cardPoints.length;
    int s = cardPoints.sublist(n - k).reduce((a, b) => a + b);
    int ans = s;
    for (int i = 0; i < k; ++i) {
      s += cardPoints[i] - cardPoints[n - k + i];
      ans = s > ans ? s : ans;
    }
    return ans;
  }
}