#2140

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Solving Questions With Brainpower

Maximize points in an exam with state transition dynamic programming by deciding whether to solve or skip each question.

array dynamic programming

Problem Statement

You are given a 0-indexed 2D integer array 'questions', where each element 'questions[i] = [pointsi, brainpoweri]' describes a question on an exam. Each question has a points value and a brainpower cost, which specifies how many questions you must skip after solving it. Your goal is to determine the maximum number of points you can earn by either solving or skipping each question.

To solve this, you must process the questions in order, making decisions to either solve or skip them. Solving a question grants points, but it prevents solving subsequent questions according to the brainpower required. Skipping a question allows you to move on to the next one. The challenge lies in balancing these options to maximize the total points earned.

Examples

Example 1

Input: questions = [[3,2],[4,3],[4,4],[2,5]]

Output: 5

The maximum points can be earned by solving questions 0 and 3.

Solve question 0: Earn 3 points, will be unable to solve the next 2 questions
Unable to solve questions 1 and 2
Solve question 3: Earn 2 points Total points earned: 3 + 2 = 5. There is no other way to earn 5 or more points.

Example 2

Input: questions = [[1,1],[2,2],[3,3],[4,4],[5,5]]

Output: 7

The maximum points can be earned by solving questions 1 and 4.

Skip question 0
Solve question 1: Earn 2 points, will be unable to solve the next 2 questions
Unable to solve questions 2 and 3
Solve question 4: Earn 5 points Total points earned: 2 + 5 = 7. There is no other way to earn 7 or more points.

Constraints

1 <= questions.length <= 105
questions[i].length == 2
1 <= pointsi, brainpoweri <= 105

Solution Approach

State Transition Dynamic Programming

This problem can be modeled with dynamic programming, where each state represents the maximum points attainable up to a given question. At each question, decide whether to solve it (and skip the next set of questions as defined by the brainpower) or skip it and proceed to the next question.

Use Memoization or Tabulation

You can implement this solution using either memoization (top-down) or tabulation (bottom-up). Both methods will maintain an array that stores the maximum points that can be earned starting from each question, allowing you to efficiently calculate the optimal decision at each step.

Greedy Decisions with Dynamic Programming

A greedy approach works alongside dynamic programming by always choosing the best possible decision at each point. You would check both options (solve or skip) and recursively compute the points for each path, storing the results to avoid recalculating the same subproblems.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexities of this solution depend on the implementation. A dynamic programming approach with memoization typically has a time complexity of O(n), where n is the number of questions, and space complexity O(n) for the memoization table. If using tabulation, space complexity can be optimized to O(1) by reusing only the last computed results.

What Interviewers Usually Probe

The candidate demonstrates a good understanding of dynamic programming with state transitions.
The candidate chooses an appropriate method (memoization or tabulation) for solving the problem efficiently.
The candidate identifies and handles edge cases, such as when it's better to skip questions based on brainpower costs.

Common Pitfalls or Variants

Common pitfalls

Not considering the effect of skipping questions and failing to track the maximum points across both choices.
Forgetting to store computed results (memoization) or using inefficient recursive calls without dynamic programming.
Improperly handling large input sizes, leading to time complexity issues without optimization.

Follow-up variants

Introduce varying brainpower costs for different questions and check if the approach remains optimal.
Modify the problem to include additional constraints such as limited time to solve questions.
Adapt the problem to allow solving multiple questions at once, complicating the decision process.

FAQ

What is the optimal approach to solving the 'Solving Questions With Brainpower' problem?

The optimal approach is to use dynamic programming, where each state represents the maximum points attainable by solving or skipping each question.

How does dynamic programming apply to 'Solving Questions With Brainpower'?

Dynamic programming helps by tracking the maximum points at each question, considering whether solving it or skipping it yields the best result.

What is the time complexity of solving 'Solving Questions With Brainpower'?

The time complexity is O(n) with memoization or tabulation, where n is the number of questions, allowing for an efficient solution.

Can the solution to 'Solving Questions With Brainpower' be optimized further?

Yes, by using tabulation with space optimization (storing only the last computed results), the space complexity can be reduced to O(1).

What are the common mistakes when solving 'Solving Questions With Brainpower'?

Common mistakes include not properly tracking the maximum points, failing to use dynamic programming effectively, and not handling large input sizes efficiently.

terminal

Solution

Solution 1: Memoization

We design a function $\textit{dfs}(i)$, which represents the maximum score that can be obtained starting from the $i$-th question. The answer is $\textit{dfs}(0)$.

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class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= len(questions):
                return 0
            p, b = questions[i]
            return max(p + dfs(i + b + 1), dfs(i + 1))

        return dfs(0)

Solution 2: Dynamic Programming

We define $f[i]$ as the maximum score that can be obtained starting from the $i$-th problem. Therefore, the answer is $f[0]$.

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class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= len(questions):
                return 0
            p, b = questions[i]
            return max(p + dfs(i + b + 1), dfs(i + 1))

        return dfs(0)

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