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1014. Best Sightseeing Pair

Description

You are given an integer array values where values[i] represents the value of the ith sightseeing spot. Two sightseeing spots i and j have a distance j - i between them.

The score of a pair (i < j) of sightseeing spots is values[i] + values[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: values = [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, values[i] + values[j] + i - j = 8 + 5 + 0 - 2 = 11

Example 2:

Input: values = [1,2]
Output: 2

 

Constraints:

  • 2 <= values.length <= 5 * 104
  • 1 <= values[i] <= 1000

Solutions

  • 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;
    }
}

  • class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& values) {
        int ans = 0, mx = values[0];
        for (int j = 1; j < values.size(); ++j) {
            ans = max(ans, values[j] - j + mx);
            mx = max(mx, values[j] + j);
        }
        return ans;
    }
};

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


  • 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
}

  • 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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