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)
1#include <vector>
2using namespace std;
3
4int maxPoints(vector<vector<int>>& questions) {
5 int n = questions.size();
6 vector<int> dp(n + 1, 0);
7 for (int i = n - 1; i >= 0; --i) {
8 int solve = questions[i][0] + (i + 1 + questions[i][1] < n ? dp[i + 1 + questions[i][1]] : 0);
9 int skip = dp[i + 1];
10 dp[i] = max(solve, skip);
11 }
12 return dp[0];
13}The C++ solution works similarly to the Python solution. It uses a vector as a dynamic array to maintain the max points from the current index to the last. Just like before, it iterates backward, calculating the best option between solving and skipping for each question. It stores results in the vector, and returns 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)
1The Python solution utilizes a recursive function dfs which includes a memoization process. The result of each recursion is cached in the memo dictionary, and recursive calls are made to either solve or skip the current question, while deciding the optimal solution based on maximum points achievable. Finally, the function returns the highest points computed from position zero.