Leetcode 1014: Best Sightseeing Pair

Input: The input consists of an array of integers, where each integer represents the value of a sightseeing spot.

Example: values = [5, 3, 6, 2, 4]

Constraints:

• 2 <= values.length <= 5 * 10^4

• 1 <= values[i] <= 1000

Output: The output should be a single integer representing the maximum score of any pair of sightseeing spots.

Example: Output: 8

Constraints:

• The result should be an integer representing the maximum score.

Goal: The goal is to find the pair of sightseeing spots that maximizes the score, which involves considering both the values and the distance between the spots.

Steps:

• Iterate through the array and for each spot, compute the score with all previous spots while keeping track of the maximum score.

• Maintain a running value for the maximum score by subtracting the distance between the spots.

Goal: Ensure the solution is efficient enough to handle the input size within the given constraints.

Steps:

• The number of sightseeing spots can be up to 50,000, so the solution must be optimized for time complexity.

• Each value in the array is between 1 and 1000.

Assumptions:

• The input will always have at least two sightseeing spots.

• Input: Input: values = [5, 3, 6, 2, 4]

• Explanation: In this case, the maximum score occurs between the first and third spots: 5 + 6 + (0 - 2) = 8.

• Input: Input: values = [1, 2]

• Explanation: For this example, the score is 1 + 2 + (0 - 1) = 2, which is the maximum score.

Approach: The optimal approach is to iterate through the array, maintaining a running maximum score based on the current spot and the previous ones. This can be done efficiently using a greedy approach.

Observations:

• We need to maximize the score, which involves both the values of the sightseeing spots and the distance between them.

• By iterating through the array while keeping track of the maximum score at each step, we can find the solution in linear time.

Steps:

• 1. Initialize a variable to store the maximum score (res).

• 2. Initialize a variable (cur) to keep track of the best score calculated so far as we iterate through the list.

• 3. Iterate through the array, updating res with the maximum score at each step.

• 4. Update cur with the best possible score for the next iteration by considering the value at the current spot and subtracting the distance from the previous spot.

Empty Inputs:

• The problem guarantees that the input will always have at least two values.

Large Inputs:

• The solution should be efficient enough to handle up to 50,000 spots in linear time.

Special Values:

• If all values are the same, the solution should still correctly calculate the maximum score based on the distances.

Constraints:

• The number of spots is at least 2, ensuring there is always a valid pair.

int maxScoreSightseeingPair(vector<int>& val) {

    int res = 0, cur = 0;

    for(int x : val) {
        res = max(res, cur + x);
        cur = max(cur, x) - 1;
    }

    return res;
}

1 : Function Declaration

int maxScoreSightseeingPair(vector<int>& val) {

Declares the function `maxScoreSightseeingPair` which takes a vector of integers `val` representing the values of sightseeing spots, and returns an integer representing the maximum score of a sightseeing pair.

2 : Initialization

    int res = 0, cur = 0;

Sets `res` to 0, as initially no score has been calculated, and `cur` to 0, which will hold the current potential score during the iteration.

3 : Loop Through Values

    for(int x : val) {

Iterates over each value `x` in the `val` vector representing the value of a sightseeing spot.

4 : Update Result

        res = max(res, cur + x);

Updates the maximum score `res` by comparing the current score `cur + x` (the sum of the current potential score and the current value) with the previous maximum score.

5 : Update Current Score

        cur = max(cur, x) - 1;

Updates the current potential score `cur` by comparing it with the value `x`. The current score is reduced by 1 as part of the problem's condition (since the second point must be after the first).

6 : Return Statement

    return res;

Returns the maximum score of a sightseeing pair, which is the result of evaluating all possible pairs.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The solution iterates through the array once, leading to a time complexity of O(n), where n is the number of sightseeing spots.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is O(1) since the solution only uses a few variables to track the maximum score and the current best score.

Leetcode 1014: Best Sightseeing Pair

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