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2140. Solving Questions With Brainpower

Description

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

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 skip each question. Solving question i will earn you pointsi points but you will be unable to solve each of the next brainpoweri 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 <= 105
  • questions[i].length == 2
  • 1 <= pointsi, brainpoweri <= 105

Solutions

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 \(\textit{dfs}(0)\).

The function \(\textit{dfs}(i)\) is calculated as follows:

  • If \(i \geq n\), it means all questions have been solved, so return \(0\);
  • Otherwise, let the score of the \(i\)-th question be \(p\), and the number of questions to skip be \(b\). Then, \(\textit{dfs}(i) = \max(p + \textit{dfs}(i + b + 1), \textit{dfs}(i + 1))\).

To avoid repeated calculations, we can use memoization by storing the values of \(\textit{dfs}(i)\) in an array \(f\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of questions.

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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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class Solution {
    private int n;
    private Long[] f;
    private int[][] questions;

    public long mostPoints(int[][] questions) {
        n = questions.length;
        f = new Long[n];
        this.questions = questions;
        return dfs(0);
    }

    private long dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        int p = questions[i][0], b = questions[i][1];
        return f[i] = Math.max(p + dfs(i + b + 1), dfs(i + 1));
    }
}

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class Solution {
public:
    long long mostPoints(vector<vector<int>>& questions) {
        int n = questions.size();
        long long f[n];
        memset(f, 0, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i) -> long long {
            if (i >= n) {
                return 0;
            }
            if (f[i]) {
                return f[i];
            }
            int p = questions[i][0], b = questions[i][1];
            return f[i] = max(p + dfs(i + b + 1), dfs(i + 1));
        };
        return dfs(0);
    }
};

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func mostPoints(questions [][]int) int64 {
    n := len(questions)
    f := make([]int64, n)
    var dfs func(int) int64
    dfs = func(i int) int64 {
        if i >= n {
            return 0
        }
        if f[i] > 0 {
            return f[i]
        }
        p, b := questions[i][0], questions[i][1]
        f[i] = max(int64(p)+dfs(i+b+1), dfs(i+1))
        return f[i]
    }
    return dfs(0)
}

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function mostPoints(questions: number[][]): number {
    const n = questions.length;
    const f = Array(n).fill(0);
    const dfs = (i: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i] > 0) {
            return f[i];
        }
        const [p, b] = questions[i];
        return (f[i] = Math.max(p + dfs(i + b + 1), dfs(i + 1)));
    };
    return dfs(0);
}

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impl Solution {
    pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
        let n = questions.len();
        let mut f = vec![-1; n];

        fn dfs(i: usize, questions: &Vec<Vec<i32>>, f: &mut Vec<i64>) -> i64 {
            if i >= questions.len() {
                return 0;
            }
            if f[i] != -1 {
                return f[i];
            }
            let p = questions[i][0] as i64;
            let b = questions[i][1] as usize;
            f[i] = (p + dfs(i + b + 1, questions, f)).max(dfs(i + 1, questions, f));
            f[i]
        }

        dfs(0, &questions, &mut f)
    }
}

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 \(f[0]\).

Considering \(f[i]\), let the score of the \(i\)-th problem be \(p\), and the number of problems to skip be \(b\). If we solve the \(i\)-th problem, then we need to solve the problem after skipping \(b\) problems, thus \(f[i] = p + f[i + b + 1]\). If we skip the \(i\)-th problem, then we start solving from the \((i + 1)\)-th problem, thus \(f[i] = f[i + 1]\). We take the maximum value of the two. The state transition equation is as follows:

\[ f[i] = \max(p + f[i + b + 1], f[i + 1]) \]

We calculate the values of \(f\) from back to front, and finally return \(f[0]\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of problems.

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class Solution:
    def mostPoints(self, questions: List[List[int]]) -> int:
        n = len(questions)
        f = [0] * (n + 1)
        for i in range(n - 1, -1, -1):
            p, b = questions[i]
            j = i + b + 1
            f[i] = max(f[i + 1], p + (0 if j > n else f[j]))
        return f[0]

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class Solution {
    public long mostPoints(int[][] questions) {
        int n = questions.length;
        long[] f = new long[n + 1];
        for (int i = n - 1; i >= 0; --i) {
            int p = questions[i][0], b = questions[i][1];
            int j = i + b + 1;
            f[i] = Math.max(f[i + 1], p + (j > n ? 0 : f[j]));
        }
        return f[0];
    }
}

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class Solution {
public:
    long long mostPoints(vector<vector<int>>& questions) {
        int n = questions.size();
        long long f[n + 1];
        memset(f, 0, sizeof(f));
        for (int i = n - 1; ~i; --i) {
            int p = questions[i][0], b = questions[i][1];
            int j = i + b + 1;
            f[i] = max(f[i + 1], p + (j > n ? 0 : f[j]));
        }
        return f[0];
    }
};

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func mostPoints(questions [][]int) int64 {
    n := len(questions)
    f := make([]int64, n+1)
    for i := n - 1; i >= 0; i-- {
        p := int64(questions[i][0])
        if j := i + questions[i][1] + 1; j <= n {
            p += f[j]
        }
        f[i] = max(f[i+1], p)
    }
    return f[0]
}

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function mostPoints(questions: number[][]): number {
    const n = questions.length;
    const f = Array(n + 1).fill(0);
    for (let i = n - 1; i >= 0; --i) {
        const [p, b] = questions[i];
        const j = i + b + 1;
        f[i] = Math.max(f[i + 1], p + (j > n ? 0 : f[j]));
    }
    return f[0];
}

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impl Solution {
    pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
        let n = questions.len();
        let mut f = vec![0; n + 1];
        for i in (0..n).rev() {
            let p = questions[i][0] as i64;
            let b = questions[i][1] as usize;
            let j = i + b + 1;
            f[i] = f[i + 1].max(p + if j > n { 0 } else { f[j] });
        }
        f[0]
    }
}

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