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Formatted question description: https://leetcode.ca/all/1014.html

1014. Best Sightseeing Pair (Medium)

Given an array A of positive integers, A[i] represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them.

The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the sum of the values of the sightseeing spots, minus the distance between them.

Return the maximum score of a pair of sightseeing spots.

 

Example 1:

Input: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11

 

Note:

  1. 2 <= A.length <= 50000
  2. 1 <= A[i] <= 1000

Companies:
Wayfair

Related Topics:
Array

Solution 1.

  • class Solution {
    public int maxScoreSightseeingPair(int[] A) {
        int maxLeftScore = A[0] + 0;
        int maxScore = 0;
        int length = A.length;
        for (int i = 1; i < length; i++) {
            int curLeftScore = A[i] + i;
            int curRightScore = A[i] - i;
            int curScore = maxLeftScore + curRightScore;
            maxScore = Math.max(maxScore, curScore);
            maxLeftScore = Math.max(maxLeftScore, curLeftScore);
        }
        return maxScore;
    }
}

############

class Solution {
    public int maxScoreSightseeingPair(int[] values) {
        int ans = 0, mx = values[0];
        for (int j = 1; j < values.length; ++j) {
            ans = Math.max(ans, values[j] - j + mx);
            mx = Math.max(mx, values[j] + j);
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/best-sightseeing-pair/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& A) {
        stack<pair<int, int>> s;
        for (int i = A.size() - 1; i >= 0; --i) {
            if (s.empty() || s.top().second < A[i] - i) s.emplace(i, A[i] - i);
        }
        int ans = INT_MIN;
        for (int i = 0; i < A.size() - 1; ++i) {
            if (s.top().first <= i) s.pop();
            ans = max(ans, A[i] + i + s.top().second);
        }
        return ans;
    }
};

  • class Solution:
    def maxScoreSightseeingPair(self, values: List[int]) -> int:
        res, mx = 0, values[0]
        for i in range(1, len(values)):
            res = max(res, values[i] - i + mx)
            mx = max(mx, values[i] + i)
        return res

############

class Solution(object):
    def shipWithinDays(self, weights, D):
        """
        :type weights: List[int]
        :type D: int
        :rtype: int
        """
        l = max(weights)
        r = sum(weights)
        # [l, r)
        while l < r:
            mid = l + (r - l) / 2
            need = 1
            cur = 0
            for w in weights:
                if cur + w > mid:
                    need += 1
                    cur = 0
                cur += w
            if need > D:
                l = mid + 1
            else:
                r = mid
        return l

  • func maxScoreSightseeingPair(values []int) (ans int) {
	for j, mx := 1, values[0]; j < len(values); j++ {
		ans = max(ans, values[j]-j+mx)
		mx = max(mx, values[j]+j)
	}
	return
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • function maxScoreSightseeingPair(values: number[]): number {
    let ans = 0;
    let mx = values[0];
    for (let j = 1; j < values.length; ++j) {
        ans = Math.max(ans, values[j] - j + mx);
        mx = Math.max(mx, values[j] + j);
    }
    return ans;
}

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