When googling I actually found the answer here. If this is homework, you should think twice before peeking :)

Algorithm::

a) Do **DFS/BFS** of the graph and keep track of the last finished vertex 'x' .

b) If there exist any mother vertex, then 'x' is one of them. Check if 'x' is a mother vertex by doing **DFS/BFS** from vertex 'x'.

Time Complexity O(n+m) + O(n+m) = **O(n+m)**

I saw the solution. I dont think we need to find SCC. Just do a DFS from a random vertex and then do the DFS from the vertex with last finish time. If there is a mother vertex then it has to be this.

**step1**. Do topological sorting of vertices of directed graph.

**step2**. Now check whether we can reach all vertices from first vertex of topologically sorted vertices in step 1.

To perform a step 2, again initialize
array *discovered[i]* to false and do dfs startin from first node of topologically sorted vertices.

If all vertices can be reached, then graph has mother vertex, and mother vertex will be the former of topologically sorted vertices.

time complexity:
step1 takes `O(n + m)`

, step 2 takes `O(n + m)`

so total `O(n+m) + O(n+m) = O(n+m)`