There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course ai first if you want to take course bi.
For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1. Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.
You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not.
Return a boolean array answer, where answer[j] is the answer to the jth query.
Example 1:
Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]] Output: [false,true] Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0. Course 0 is not a prerequisite of course 1, but the opposite is true. Example 2:
Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]] Output: [false,false] Explanation: There are no prerequisites, and each course is independent. Example 3:
Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]] Output: [true,true]
##code in a c++ #include <bits/stdc++.h> using namespace std;
static auto init = { ios::sync_with_stdio(false); cin.tie(nullptr); return nullptr; }();
class Solution { public: vector checkIfPrerequisite(int n, vector<vector>& prerequisites, vector<vector>& queries) { vector<vector> adj(n); for (auto &p : prerequisites) adj[p[0]].push_back(p[1]); vector<bitset<101>> memo(n); vector visited(n, false);
function<void(int)> dfs = [&](int s) {
visited[s] = true;
for (auto t : adj[s]) {
if (!visited[t]) dfs(t);
memo[s] |= memo[t];
memo[s].set(t);
}
};
for (int i = 0; i < n; ++i) {
if (!visited[i]) dfs(i);
}
vector<bool> ans;
ans.reserve(queries.size());
for (auto &q : queries) {
ans.push_back(memo[q[0]][q[1]]);
}
return ans;
}
};