LeetCode 1014. Best Sightseeing Pair Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Enumeration

We can enumerate j from left to right while maintaining the maximum value of values[i] + i for elements to the left of j, denoted as mx. For each j, the maximum score is mx + values[j] - j. The answer is the maximum of these maximum scores for all positions.

The time complexity is O(n), where n is the length of the array values. The space complexity is O(1).

PythonJavaC++GoTypeScriptRust

class Solution: def maxScoreSightseeingPair(self, values: List[int]) -> int: ans = mx = 0 for j, x in enumerate(values): ans = max(ans, mx + x - j) mx = max(mx, x + j) return ans(code-box)

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

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

func maxScoreSightseeingPair(values []int) (ans int) { mx := 0 for j, x := range values { ans = max(ans, mx+x-j) mx = max(mx, x+j) } return }(code-box)

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

impl Solution { pub fn max_score_sightseeing_pair(values: Vec<i32>) -> i32 { let mut ans = 0; let mut mx = 0; for (j, &x) in values.iter().enumerate() { ans = ans.max(mx + x - j as i32); mx = mx.max(x + j as i32); } ans } }(code-box)