Detect cycle in a directed graph

// { Driver Code Starts
#include <bits/stdc++.h>
using namespace std;


 // } Driver Code Ends
/*  Function to check if the given graph contains cycle
*   V: number of vertices
*   adj[]: representation of graph
*/
bool isCyclic_util(vector<int> adj[], vector<bool> visited, int curr)
{
    if(visited[curr]==true)
        return true;
    
    visited[curr] = true;
    bool FLAG = false;
    for(int i=0;i<adj[curr].size();++i)
    {
        FLAG = isCyclic_util(adj, visited, adj[curr][i]);
        if(FLAG==true)
            return true;
    }
    return false;
}

bool isCyclic(int V, vector<int> adj[])
{
   vector<bool> visited(V,false);
   bool FLAG = false;
   for(int i=0;i<V;++i)
   {
           visited[i] = true;
           for(int j=0;j<adj[i].size();++j)
           {
               FLAG = isCyclic_util(adj,visited,adj[i][j]);
               if(FLAG==true)
                   return true;
           }
           visited[i] = false;
   }
   return false;
}

// { Driver Code Starts.

int main() {
	
	int t;
	cin >> t;
	
	while(t--){
	    
	    int v, e;
	    cin >> v >> e;
	    
	    vector<int> adj[v];
	    
	    for(int i =0;i<e;i++){
	        int u, v;
	        cin >> u >> v;
	        adj[u].push_back(v);
	    }
	    
	    cout << isCyclic(v, adj) << endl;
	    
	}
	
	return 0;
}  // } Driver Code Ends

checkout this, simple DFS solution with memoization, this will pass all test cases
pardon for golang code, but it is understandable
https://gist.github.com/rahulkushwaha12/775da82fa280effce68991d92ea9c544

sir, after line 23, is it needed to put visited[curr]==false or it takes directly false. can you explain this sir..???

Same doubt as we are taking completely new visited array we have to make all the elements false after every iteration of the for loop.
Here is the psuedocode
.............
...........
for .....{
visited[] = false*len(visited);
...................
..............
}

What will be the time complexity?

What will be the time complexity?

Average / In General : O(V+E)
Words Case : O(V^{2}) when every node is connected to all the other nodes.