# Course Schedule IV in C++

C++Server Side ProgrammingProgramming

Suppose there are a total of n courses we can to take, the courses are labeled from 0 to n-1.

Some courses may have direct prerequisites, as example, to take course 0 we have first to take course 1, which is expressed as a pair: [1,0].

So, if we have a number of courses n, a list of direct prerequisite pairs and a list of queries pairs.

You should find the answer for each queries[i] whether the course queries[i] is a prerequisite of the course queries[i] or not. Finally, we have to return a list of boolean, the answers to the given queries.

We have to keep in mind that 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.

So, if the input is like n = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]], then the output will be [true,true]

To solve this, we will follow these steps −

• N := 110

• Define an array ret

• Define one map in

• for each element it in v, do

• insert it at the end of graph[it]

• (increase in[it] by 1)

• Define one queue q

• for initialize i := 0, when i < n, update (increase i by 1), do −

• if in[i] is same as 0, then −

• insert i into q

• Define one map idx

• for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −

• sz := size of q

• while sz is not 0, decrease sz in each iteration, do −

• node := first element of q

• delete element from q

• for each element it in graph[node]

• (decrease in[it] by 1)

• for each element x in c[node], do

• insert x into c[it]

• insert node into c[it]

• if in[it] is same as 0, then −

• insert it into q

• for each element it in x, do

• insert (frequency of it in c[it]) at the end of ret

• return ret

## Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<bool> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
const int N = 110;
class Solution {
public:
vector <int> graph[N];
map <int, set <int>> c;
vector<bool> checkIfPrerequisite(int n, vector<vector<int>>& v, vector<vector<int>>& x) {
vector<bool> ret;
map<int, int> in;
for (auto& it : v) {
graph[it].push_back(it);
in[it]++;
}
queue<int> q;
for (int i = 0; i < n; i++) {
if (in[i] == 0)
q.push(i);
}
map<int, int> idx;
for (int lvl = 1; !q.empty(); lvl++) {
int sz = q.size();
while (sz--) {
int node = q.front();
q.pop();
for (auto& it : graph[node]) {
in[it]--;
for (auto& x : c[node])
c[it].insert(x);
c[it].insert(node);
if (in[it] == 0) {
q.push(it);
}
}
}
}
for (auto& it : x) {
ret.push_back(c[it].count(it));
}
return ret;
}
};
main(){
Solution ob;
vector<vector<int>> prerequisites = {{1,2},{1,0},{2,0}}, queries = {{1,0},{1,2}};
print_vector(ob.checkIfPrerequisite(3, prerequisites, queries));
}

## Input

3, {{1,2},{1,0},{2,0}}, {{1,0},{1,2}}

## Output

[1, 1, ]