1014

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.

// OJ: https://leetcode.com/problems/best-sightseeing-pair/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/best-sightseeing-pair/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& A) {
        int maxVal = A[0], ans = INT_MIN;
        for (int i = 1; i < A.size(); ++i) {
            ans = max(ans, maxVal + A[i] - i);
            maxVal = max(maxVal, A[i] + i);
        }
        return ans;
    }
};