2140. Solving Questions With Brainpower

2140. Solving Questions With Brainpower

Tag: Dynamic Programming

Difficulty: Medium

You are given a 0-indexed 2D integer array questions where questions[i] = [points_i, brainpower_i].

The array describes the questions of an exam, where you have to process the questions in order (i.e., starting from question 0) and make a decision whether to solve or to skip each question. Solving question i will earn you points_i points but you will be unable to solve each of the next brainpower_i questions. If you skip question i, you get to make the decision on the next question.

• For example, given questions = [[3, 2], [4, 3], [4, 4], [2, 5]]:
  • If question 0 is solved, you will earn 3 points but you will be unable to solve questions 1 and 2.
  • If instead, question 0 is skipped and question 1 is solved, you will earn 4 points but you will be unable to solve questions 2 and 3.

Return the maximum points you can earn for the exam.

---

Example 1:

Input: questions = [[3,2],[4,3],[4,4],[2,5]]
Output: 5
Explanation: 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
Explanation: 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 <= 10^5
• questions[i].length == 2
• 1 <= points<sub>i</sub>, brainpower<sub>i</sub> <= 10^5

---

The Framework

Top-Down Dynamic Programming

class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        n = len(questions)

        @lru_cache(None)
        def dp(curr):
            # Base case
            if curr >= n:
                return 0
            
            points, brainpower = questions[curr][0], questions[curr][1]

            # Recurrence relation: to solve or to skip the current question?
            # Skip: move on to the next question
            skip = dp(curr + 1)
            # Solve: collect the point of current question, move on to the next possible question from brainpower
            solve = dp(curr + brainpower + 1) + points
            # What is the best decision though?
            return max(skip, solve)

        return dp(0)

class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        n = len(questions)
        memo = collections.defaultdict(int)

        def dp(curr):
            # Base case
            if curr >= n:
                return 0
            
            points, brainpower = questions[curr][0], questions[curr][1]

            if curr in memo:
                return memo[curr]

            # Recurrence relation: to solve or to skip the current question?
            # Skip: move on to the next question
            skip = dp(curr + 1)
            # Solve: collect the point of current question, move on to the next possible question from brainpower
            solve = dp(curr + brainpower + 1) + points
            # What is the best decision though?
            memo[curr] = max(skip, solve)
            return memo[curr]

        return dp(0)

Bottom-Up Dynamic Programming

class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        n = len(questions)
        dp = [0] * (n + 1)

        for curr in range(n - 1, -1, -1):
            points, brainpowers = questions[curr][0], questions[curr][1]
            skip, pick = 0, points
            if curr + brainpowers + 1 < n:
                pick += dp[curr + brainpowers + 1]
            if curr + 1 < n:
                skip += dp[curr + 1]
            dp[curr] = max(skip, pick)
        
        return dp[0]

Algorithm

1. Initialize an array dp of size n, set dp[n - 1] = questions[n - 1][0].

2. Iterate backward over index i from n - 2:

• If we skip question i, we have dp[i] = dp[i + 1].
• If we solve question i, we have dp[i] = questions[i][0] + dp[i + questions[i][1] + 1].

3. Update dp[i] as the larger one.

4. Return dp[0] once we finish the iteration.

class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        n = len(questions)
        dp = [0] * n
        dp[-1] = questions[-1][0]
        
        for curr in range(n - 2, -1, -1):
            dp[curr], skip = questions[curr][0], questions[curr][1]
            
            if curr + skip + 1 < n:
                dp[curr] += dp[curr + skip + 1]
            
            dp[curr] = max(dp[curr], dp[curr + 1])
        
        return dp[0]