What is the time complexity of Maximum Points Tourist Can Earn?

The optimal solution has a time complexity of O(n^2*k).

What data structures are used to solve Maximum Points Tourist Can Earn?

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

What are the key insights for solving Maximum Points Tourist Can Earn?

Insight 1: Dynamic programming is suitable for this optimization problem because the optimal solution for `k` days can be built upon the optimal solutions for `k-1` days. Insight 2: The state of the dynamic programming solution depends on the current day and the city the tourist is in.

Which companies ask LeetCode 3332. Maximum Points Tourist Can Earn in interviews?

This problem is frequently asked at DE Shaw.

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Maximum Points Tourist Can Earn - LeetCode 3332 Solution

Maximum Points Tourist Can Earn - Complete Solution Guide

Maximum Points Tourist Can Earn is LeetCode problem 3332, 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 two integers, n and k , along with two 2D integer arrays, stayScore and travelScore . A tourist is visiting a country with n cities, where each city is directly connected to every other city. The tourist's journey consists of exactly k 0-indexed days, and they can choose any city as their starting point. Each day, the tourist has two choices: Stay in the current city : If the tourist stays in their current city curr during day i , they will earn stayScore[i][curr] points. Move to a

Detailed Explanation

The problem asks us to find the maximum points a tourist can earn while visiting `n` cities over `k` days. The tourist can either stay in the current city or move to another city each day. Staying in a city earns points based on `stayScore`, while moving earns points based on `travelScore`. The goal is to maximize the total points earned over the `k` days, considering all possible starting cities and travel/stay decisions.

Solution Approach

The solution uses dynamic programming to calculate the maximum points achievable at each city for each day. It maintains a 1D array `dp` of size `n`, where `dp[j]` represents the maximum points achievable at city `j` on the current day. The code iterates through the days from 1 to `k-1`, updating the `dp` array in each iteration. For each city `j`, it considers two options: staying in the current city or moving from another city. The `dp[j]` is updated with the maximum of these two options.

Step-by-Step Algorithm

Step 1: Initialize `dp` array of size `n` with 0s. This array will store the maximum score achievable at each city on the current day.
Step 2: For the first day (day 0), calculate the initial `dp` values for each city. For each city `j`, consider staying in that city (using `stayScore[0][j]`) and moving from any other city `s` (using `travelScore[s][j]`). The `dp[j]` will be the maximum of the stay score and the maximum travel score.
Step 3: Iterate from day 1 to `k-1`. For each day `i`, create a new `dp` array called `new_dp`. This is needed to avoid overwriting previous day's values before all calculations for current day are finished.
Step 4: For each city `j` on day `i`, calculate the score if the tourist stays at city `j` (`dp[j] + stayScore[i][j]`).
Step 5: Calculate the maximum score if the tourist moves to city `j` from another city `c`. This involves iterating through all other cities `c` and calculating `dp[c] + travelScore[c][j]`, and selecting the maximum value among all these possible moves.
Step 6: Update `new_dp[j]` with the maximum score achievable between staying at city `j` and moving to city `j` from other cities. In particular, `new_dp[j] = max(score_if_stay, max_score_if_move)`.
Step 7: After calculating `new_dp` for all cities on day `i`, update `dp` with `new_dp`. This prepares for the next iteration.
Step 8: After iterating through all `k` days, find the maximum value in the `dp` array, which represents the maximum points the tourist can earn.

Key Insights

Insight 1: Dynamic programming is suitable for this optimization problem because the optimal solution for `k` days can be built upon the optimal solutions for `k-1` days.
Insight 2: The state of the dynamic programming solution depends on the current day and the city the tourist is in.
Insight 3: Since the number of cities (n) and days (k) are relatively small (<= 200), an O(n^2 * k) time complexity solution is acceptable.

Complexity Analysis

Time Complexity: O(n^2*k)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: DE Shaw.

Maximum Points Tourist Can Earn | LeetCode 3332

Medium

Array Dynamic Programming Matrix

Time: O(n^2*k)Space: O(n)

Solution Code

Available in:

class Solution {
    public int maxScore(int n, int k, int[][] stayScore, int[][] travelScore) {
        int[] dp = new int[n];

        // Day 0
        for (int j = 0; j < n; j++) {
            int score_stay = stayScore[0][j];
            int max_score_travel = 0;
            for (int s = 0; s < n; s++) {
                max_score_travel = Math.max(max_score_travel, travelScore[s][j]);
            }
            dp[j] = Math.max(score_stay, max_score_travel);
        }

        // Day 1 to k-1
        for (int i = 1; i < k; i++) {
            int[] new_dp = new int[n];
            for (int j = 0; j < n; j++) {
                int score_if_stay = dp[j] + stayScore[i][j];
                
                int max_score_if_move = 0;
                for (int c = 0; c < n; c++) {
                    int score = dp[c] + travelScore[c][j];
                    max_score_if_move = Math.max(max_score_if_move, score);
                }
                
                new_dp[j] = Math.max(score_if_stay, max_score_if_move);
            }
            dp = new_dp;
            
        }

        int max = 0;
        if (dp.length == 0) return 0;
        for (int val : dp) {
            max = Math.max(max, val);
        }
        return max;
    }
}

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Problem Info

Description

Problem 3332: Maximum Points Tourist Can Earn - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3332

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

8 steps

1
Step 1: Initialize `dp` array of size `n` with 0s. This array will store the maximum score achievable at each city on the current day.
2
Step 2: For the first day (day 0), calculate the initial `dp` values for each city. For each city `j`, consider staying in that city (using `stayScore[0][j]`) and moving from any other city `s` (using `travelScore[s][j]`). The `dp[j]` will be the maximum of the stay score and the maximum travel score.
3
Step 3: Iterate from day 1 to `k-1`. For each day `i`, create a new `dp` array called `new_dp`. This is needed to avoid overwriting previous day's values before all calculations for current day are finished.
4
Step 4: For each city `j` on day `i`, calculate the score if the tourist stays at city `j` (`dp[j] + stayScore[i][j]`).
5
Step 5: Calculate the maximum score if the tourist moves to city `j` from another city `c`. This involves iterating through all other cities `c` and calculating `dp[c] + travelScore[c][j]`, and selecting the maximum value among all these possible moves.
6
Step 6: Update `new_dp[j]` with the maximum score achievable between staying at city `j` and moving to city `j` from other cities. In particular, `new_dp[j] = max(score_if_stay, max_score_if_move)`.
7
Step 7: After calculating `new_dp` for all cities on day `i`, update `dp` with `new_dp`. This prepares for the next iteration.
8
Step 8: After iterating through all `k` days, find the maximum value in the `dp` array, which represents the maximum points the tourist can earn.

💡

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^2*k)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the nested loops used to iterate through the days, cities, and previous cities to calculate the maximum scores. The space complexity is dominated by the 1D array `dp` used to store intermediate results.

📖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

Recursion with Memoization

A recursive approach can be used with memoization to avoid redundant calculations. The recursive function would take the current day and current city as input and return the maximum score achievable from that state. A 2D array could be used to store the results of the recursive calls.

Time: O(n^2 * k)Space: O(n * k)

Trade-offs: The recursive approach with memoization has the same time complexity as the iterative dynamic programming approach, but it may have a higher space complexity due to the recursion stack and the memoization table. The iterative approach is generally preferred for its better space efficiency.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for this optimization problem because the optimal solution for `k` days can be built upon the optimal solutions for `k-1` days.
2Insight 2: The state of the dynamic programming solution depends on the current day and the city the tourist is in.
3Insight 3: Since the number of cities (n) and days (k) are relatively small (<= 200), an O(n^2 * k) time complexity solution is acceptable.

📚

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: `n = 1`, meaning only one city. The code correctly handles this case as it iterates through the single city. The travel score will not be used, and the solution will involve just staying at the city `k` times.
Edge case 2: `k = 1`, meaning only one day. The code calculates the maximum score between staying in the current city and moving from other cities and correctly return the overall max score.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all possible starting cities. The code should implicitly consider all possible starting cities during the first day's calculation by calculating the max scores for all the cities as starting points.

✗Common mistake 2: Not correctly handling the transition between days. The `dp` array must be updated correctly at each day, taking into account both staying and moving to the current city.

✅

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: Resource allocation in a network where nodes represent resources, and the goal is to maximize resource utilization over a period of time. The `stayScore` could represent the benefit of using the resource locally, and `travelScore` the cost/benefit of transferring resources between nodes.

🏦Real-world scenario 2: Optimizing a delivery route. Cities are locations to deliver to, `stayScore` represents profit from delivering to a place, and `travelScore` the costs of going from one place to another. The goal is to maximize profit during a set number of stops.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 1, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 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 "Maximum Points Tourist Can Earn"?

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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