Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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.
questions = [[3, 2], [4, 3], [4, 4], [2, 5]]:
0 is solved, you will earn 3 points but you will be unable to solve questions 1 and 2.0 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 <= 105questions[i].length == 21 <= pointsi, brainpoweri <= 105Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming
[[3,2],[4,3],[4,4],[2,5]]
[[1,1],[2,2],[3,3],[4,4],[5,5]]
house-robber)frog-jump)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2140: Solving Questions With Brainpower
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));
}
}
// Accepted solution for LeetCode #2140: Solving Questions With Brainpower
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)
}
# Accepted solution for LeetCode #2140: Solving Questions With Brainpower
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)
// Accepted solution for LeetCode #2140: Solving Questions With Brainpower
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)
}
}
// Accepted solution for LeetCode #2140: Solving Questions With Brainpower
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);
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.