Sponsored
Sponsored
This approach leverages dynamic programming to solve the problem. You create a DP array where each element dp[i] represents the maximum points that can be earned starting from question i to the last question. You decide whether to solve or skip a question based on the potential points you can earn. The solution is filled backward, starting from the last question.
Time Complexity: O(n)
Space Complexity: O(n)
1def maxPoints(questions):
2 n = len(questions)
3 dp = [0] * (n + 1)
4 for i in range(n - 1, -1, -1):
5 solve = questions[i][0] + dp[min(n, i + 1 + questions[i][1])]
6 skip = dp[i + 1]
7 dp[i] = max(solve, skip)
8 return dp[0]The code initializes a dp array with zeros of size n+1 where n is the number of questions. It iterates over the questions in reverse order, calculating the maximum possible points by considering either solving or skipping the current question. Finally, it returns the result stored in dp[0] which represents the maximum points from the first question.
This approach uses recursion with memoization to avoid recomputing results for overlapping subproblems. It recursively decides the maximum points by considering both solving and skipping options for each question, and stores results in a memoization array for reuse. This provides an efficient way to solve the problem since it avoids redundant calculations.
Time Complexity: O(n)
Space Complexity: O(n)
1#include <unordered_map>
using namespace std;
int dfs(vector<vector<int>>& questions, int i, unordered_map<int, int>& memo) {
if (i >= questions.size()) return 0;
if (memo.count(i)) return memo[i];
int solve = questions[i][0] + dfs(questions, i + 1 + questions[i][1], memo);
int skip = dfs(questions, i + 1, memo);
memo[i] = max(solve, skip);
return memo[i];
}
int maxPoints(vector<vector<int>>& questions) {
unordered_map<int, int> memo;
return dfs(questions, 0, memo);
}In the C++ code, a recursive function dfs with memoization calculates the maximum points starting from each question. Memoization is implemented using a hashmap for caching results, which helps mitigate redundant operations by storing and reusing prior computations. The code evaluates points from solving or skipping and consistently chooses the option providing maximum points, finally returning results starting from the first question.