What is the time complexity of Best Sightseeing Pair?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Best Sightseeing Pair?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Best Sightseeing Pair?

Insight 1: The core idea is to decompose the score function `values[i] + values[j] + i - j` into two parts: `(values[i] + i)` and `(values[j] - j)`. This allows us to precompute the maximum value of `(values[i] + i)` for all `i < j`. Insight 2: We can use a single pass through the array to keep track of the maximum value of `values[i] + i` encountered so far and calculate the maximum score at each step.

Which companies ask LeetCode 1014. Best Sightseeing Pair in interviews?

This problem is frequently asked at Wayfair.

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Best Sightseeing Pair - LeetCode 1014 Solution

Q: How to solve Best Sightseeing Pair?

The provided solution uses a single pass through the array `values` to calculate the maximum score. It maintains a variable `max_so_far` that stores the maximum value of `values[i] + i` encountered so far. For each element `values[j]`, it calculates the score `max_so_far + values[j] - j` and updates the `max_score` variable if the calculated score is greater than the current `max_score`. Then, it updates `max_so_far` to `max(max_so_far, values[j] + j)` to consider the current spot for future calculations. This approach effectively finds the optimal pair `(i, j)` by maximizing `values[i] + i` for each potential `j`.

Q: Which companies ask LeetCode 1014. Best Sightseeing Pair in interviews?

This problem is frequently asked at Wayfair.

Best Sightseeing Pair - Complete Solution Guide

Best Sightseeing Pair is LeetCode problem 1014, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array values where values[i] represents the value of the i th 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 -

Detailed Explanation

The problem asks us to find the maximum score of a pair of sightseeing spots in a given array of values. The score is calculated as `values[i] + values[j] + i - j`, where `i` and `j` are the indices of the two sightseeing spots, and `i < j`. The array `values` contains the value of each sightseeing spot. The goal is to maximize the score across all possible pairs of `i` and `j` that satisfy `i < j`.

Solution Approach

The provided solution uses a single pass through the array `values` to calculate the maximum score. It maintains a variable `max_so_far` that stores the maximum value of `values[i] + i` encountered so far. For each element `values[j]`, it calculates the score `max_so_far + values[j] - j` and updates the `max_score` variable if the calculated score is greater than the current `max_score`. Then, it updates `max_so_far` to `max(max_so_far, values[j] + j)` to consider the current spot for future calculations. This approach effectively finds the optimal pair `(i, j)` by maximizing `values[i] + i` for each potential `j`.

Step-by-Step Algorithm

Step 1: Initialize `max_so_far` with `values[0]`. This will store the maximum value of `values[i] + i` for all `i` considered so far.
Step 2: Initialize `max_score` to 0. This will store the maximum score encountered so far.
Step 3: Iterate through the `values` array from index `j = 1` to `len(values) - 1`.
Step 4: In each iteration, calculate the score `max_so_far + values[j] - j`.
Step 5: Update `max_score` by taking the maximum of the current `max_score` and the calculated score from Step 4: `max_score = max(max_score, max_so_far + values[j] - j)`.
Step 6: Update `max_so_far` by taking the maximum of the current `max_so_far` and `values[j] + j`: `max_so_far = max(max_so_far, values[j] + j)`.
Step 7: After the loop finishes, return `max_score`.

Key Insights

Insight 1: The core idea is to decompose the score function `values[i] + values[j] + i - j` into two parts: `(values[i] + i)` and `(values[j] - j)`. This allows us to precompute the maximum value of `(values[i] + i)` for all `i < j`.
Insight 2: We can use a single pass through the array to keep track of the maximum value of `values[i] + i` encountered so far and calculate the maximum score at each step.
Insight 3: The problem can be solved in O(n) time and O(1) space by iteratively updating the `max_so_far` variable.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Wayfair.

Best Sightseeing Pair | LeetCode 1014

Medium

Array Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxScoreSightseeingPair(int[] values) {
        int max_so_far = values[0];
        int max_score = 0;

        for (int j = 1; j < values.length; j++) {
            // The score is (values[i] + i) + (values[j] - j).
            // max_so_far keeps track of the maximum (values[i] + i) for i < j.
            max_score = Math.max(max_score, max_so_far + values[j] - j);
            // Update max_so_far to include the current j for future calculations.
            max_so_far = Math.max(max_so_far, values[j] + j);
        }

        return max_score;
    }
}

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Problem Info

Description

Problem 1014: Best Sightseeing Pair - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

Wayfair

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1014

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `max_so_far` with `values[0]`. This will store the maximum value of `values[i] + i` for all `i` considered so far.
2
Step 2: Initialize `max_score` to 0. This will store the maximum score encountered so far.
3
Step 3: Iterate through the `values` array from index `j = 1` to `len(values) - 1`.
4
Step 4: In each iteration, calculate the score `max_so_far + values[j] - j`.
5
Step 5: Update `max_score` by taking the maximum of the current `max_score` and the calculated score from Step 4: `max_score = max(max_score, max_so_far + values[j] - j)`.
6
Step 6: Update `max_so_far` by taking the maximum of the current `max_so_far` and `values[j] + j`: `max_so_far = max(max_so_far, values[j] + j)`.
7
Step 7: After the loop finishes, return `max_score`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm iterates through the input array once. The space complexity is constant because the algorithm uses a fixed number of variables to store intermediate results, and the memory usage doesn't depend on the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible pairs (i, j) with i < j and calculate the score for each pair. Then, find the maximum score among all pairs.

Time: O(n^2)Space: O(1)

Trade-offs: The brute force approach is simple to understand but has a quadratic time complexity, making it inefficient for large input arrays. The given O(n) solution is much better for larger input sizes.

✨

Key Insights

1Insight 1: The core idea is to decompose the score function `values[i] + values[j] + i - j` into two parts: `(values[i] + i)` and `(values[j] - j)`. This allows us to precompute the maximum value of `(values[i] + i)` for all `i < j`.
2Insight 2: We can use a single pass through the array to keep track of the maximum value of `values[i] + i` encountered so far and calculate the maximum score at each step.
3Insight 3: The problem can be solved in O(n) time and O(1) space by iteratively updating the `max_so_far` variable.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Input array with only two elements. The solution correctly calculates the score and returns the maximum.
Edge case 2: Input array where all values are 1. The solution still correctly finds the maximum score according to the defined formula.
Edge case 3: Input array with decreasing values. The algorithm iterates through the array and computes the max_so_far and updates the max_score accordingly, ensuring it finds the optimal solution in any case.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the formula `values[i] + values[j] + i - j` and how to efficiently compute the maximum score.

✗Common mistake 2: Using nested loops to calculate the score for all pairs, resulting in O(n^2) time complexity instead of the optimal O(n) time complexity.

✗Common mistake 3: Failing to update `max_so_far` correctly, leading to incorrect calculation of the maximum score.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing the selection of two advertisements to maximize their combined impact, considering their individual values and the delay (distance) between showing them.

🏦Real-world scenario 2: Finding the best pair of trading opportunities in a financial market, where values represent potential profits, and the distance represents the time delay between trades.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Best Sightseeing Pair"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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