<p>There are a total of <code>numCourses</code> courses you have to take, labeled from <code>0</code> to <code>numCourses - 1</code>. You are given an array <code>prerequisites</code> where <code>prerequisites[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that you <strong>must</strong> take course <code>b<sub>i</sub></code> first if you want to take course <code>a<sub>i</sub></code>.</p>

<ul>
	<li>For example, the pair <code>[0, 1]</code>, indicates that to take course <code>0</code> you have to first take course <code>1</code>.</li>
</ul>

<p>Return <code>true</code> if you can finish all courses. Otherwise, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> numCourses = 2, prerequisites = [[1,0]]
<strong>Output:</strong> true
<strong>Explanation:</strong> There are a total of 2 courses to take. 
To take course 1 you should have finished course 0. So it is possible.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> numCourses = 2, prerequisites = [[1,0],[0,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> There are a total of 2 courses to take. 
To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= numCourses &lt;= 2000</code></li>
	<li><code>0 &lt;= prerequisites.length &lt;= 5000</code></li>
	<li><code>prerequisites[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; numCourses</code></li>
	<li>All the pairs prerequisites[i] are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    bool canFinish(int V, vector<vector<int>>& pre_req) {
        vector<int>adj[V];
        //making adjacency list;
        for(auto i: pre_req){
            adj[i[1]].push_back(i[0]);
        }
      //we will solve this question using topological sort;
      //writing code for toposort;
      //we will take the concept of indegree, if any node which is having indegree 0, we will push it to the queue;
      //and correspondingly , we will reduce the indegree of its neighbour;
      vector<int>indegree(V,0);
      //traversing;
      for(int i=0; i<V; i++){
          for(auto i: adj[i] ){
              indegree[i]++;
          }
      }
      //now making a queue;
      queue<int>q;
      for(int i=0; i<V; i++){
          if(indegree[i] == 0){
              q.push(i);
          }
      }
      //now making a vector for toposort;
      vector<int>topo;
      while(!q.empty()){
          int temp = q.front();
          q.pop();
          topo.push_back(temp);
          //now reducing the indegree;
          for(auto i: adj[temp]){
              indegree[i]--;
              if(indegree[i] == 0)q.push(i);
          }
      }
      //at last checking the condition if a cycle exist;
      if(topo.size() == V){
          return true;
      }
      return false;
        
    }
};
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