[leetcode] 1423. Maximum Points You Can Obtain from Cards

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

카드를 좌, 우에서 한장씩 k 번의 갯수만큼 획득하는데, 획득할 수 있는 최대 point 를 계산하는 문제이다. 

단순 top down dp 로 풀면 TLE 에 걸린다.

 

잘 생각해보면 좌, 우에서 일정부분을 뺀 값이 최대값이 된다면, 나머지 부분은 최소값이 되어야 한다는 것을 알 수 있다.

즉 전체 array sum 에서 가운데 subarray 는 최소값이 되어야 하고, 이 subarray 의 min 값은 $O(N)$ 안에 구할 수가 있다.

 

class Solution {
    public int maxScore(int[] cardPoints, int k) {

		int n = cardPoints.length;
		int totalSum = 0;
		int sum = 0;
		int min = Integer.MAX_VALUE;
		int windowSize = n - k;

		for (int i = 0; i < cardPoints.length; i++) {
			totalSum += cardPoints[i];
		}

		for (int i = 0; i < windowSize; i++) {
			sum += cardPoints[i];
		}

		min = Math.min(min, sum);

		if (windowSize > 0) {
			for (int right = windowSize; right < cardPoints.length; right++) {
				int left = right - windowSize + 1;

				sum -= cardPoints[left-1];
				sum += cardPoints[right];

				min = Math.min(min, sum);
			}
		}

		return totalSum - min;
	}
}