How do you solve Best Sightseeing Pair on LeetCode?

Best Sightseeing Pair (LeetCode #1014) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The score formula can be rearranged to maximize the sum of values and minimize the distance.

What is the brute force solution for Best Sightseeing Pair?

The brute force approach for Best Sightseeing Pair is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of sightseeing spots to calculate their scores. This ensures we find the maximum score, but it can be slow for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Best Sightseeing Pair?

Best Sightseeing Pair uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Array, Dynamic Programming.

What are the interview tips for Best Sightseeing Pair?

Always think about optimizing nested loops when possible. Consider how to break down the problem and track necessary values during iteration. Practice explaining your thought process clearly, as it helps solidify your understanding.

What are common mistakes when solving Best Sightseeing Pair?

Not considering the impact of the distance in the score calculation. Failing to update the best value as we iterate.

#1014

Best Sightseeing Pair

Medium

ArrayDynamic ProgrammingArrayDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking all pairs, we can keep track of the best score we can achieve for the first sightseeing spot as we iterate through the list. This allows us to calculate the maximum score in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the maximum score and another to keep track of the best value of the first sightseeing spot adjusted for its index.
2Step 2: Iterate through the array, updating the best value as we go.
3Step 3: For each index j, calculate the score using the best value from previous indices and update the maximum score.

solution.py7 lines

1def maxScoreSightseeingPair(values):
2    max_score = 0
3    best_value = values[0]
4    for j in range(1, len(values)):
5        max_score = max(max_score, best_value + values[j] - j)
6        best_value = max(best_value, values[j] + j)
7    return max_score

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array. The space complexity is O(1) since we are using a constant amount of extra space.

1The score formula can be rearranged to maximize the sum of values and minimize the distance.
2Keeping track of the best previous value allows for a single pass solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.