How do you solve Maximum Points Tourist Can Earn on LeetCode?

Maximum Points Tourist Can Earn (LeetCode #3332) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^2 * k) time complexity and O(n * k) space complexity. Key insight: Dynamic programming helps optimize the exploration of paths by storing intermediate results.

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

The optimal solution for Maximum Points Tourist Can Earn runs in O(n^2 * k) time and uses O(n * k) space. This complexity arises because we fill a DP table of size k x n, where each cell requires checking all previous cities.

What is the brute force solution for Maximum Points Tourist Can Earn?

The brute force approach for Maximum Points Tourist Can Earn is Brute Force, with O(n^k) time complexity and O(k) space complexity. The brute-force approach tries all possible combinations of staying and traveling across the cities for each day. This method is straightforward but inefficient as it explores every possible path. The optimal Optimal Solution approach improves this to O(n^2 * k).

What algorithm or data structure is used to solve Maximum Points Tourist Can Earn?

Maximum Points Tourist Can Earn uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Maximum Points Tourist Can Earn?

Always explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Be ready to discuss the trade-offs between time and space complexity. Practice explaining your code as you write it, as this simulates the interview environment.

What are common mistakes when solving Maximum Points Tourist Can Earn?

Not considering all possible paths, especially when moving between cities. Failing to initialize the DP table correctly, leading to incorrect results.

#3332

Maximum Points Tourist Can Earn

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^k)	O(n^2 * k)
Space	O(k)	O(n * k)

💡

Intuition

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

The optimal solution uses dynamic programming to build up the maximum points achievable day by day, considering both staying and traveling options. This significantly reduces the number of calculations by storing intermediate results.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[day][city] represents the maximum points achievable on 'day' while in 'city'.
2Step 2: For each day, calculate the maximum points for each city by considering both staying and moving to other cities.
3Step 3: Return the maximum value from the last day across all cities.

solution.py8 lines

1def maxPoints(n, k, stayScore, travelScore):
2    dp = [[0] * n for _ in range(k)]
3    for city in range(n):
4        dp[0][city] = stayScore[0][city]
5    for day in range(1, k):
6        for city in range(n):
7            dp[day][city] = stayScore[day][city] + max(dp[day - 1][c] + travelScore[c][city] for c in range(n))
8    return max(dp[k - 1])

ℹ

Complexity note: This complexity arises because we fill a DP table of size k x n, where each cell requires checking all previous cities.

1Dynamic programming helps optimize the exploration of paths by storing intermediate results.
2Understanding the structure of the problem allows for breaking it down into manageable subproblems.

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