Find a Mother Vertex in a Graph
Last Updated :
23 Jul, 2025
Write a function to find a mother vertex in the graph.
What is a Mother Vertex?
A mother vertex in a graph G = (V, E) is a vertex v such that all other vertices in G can be reached by a path from v
Example:
Input: Graph as shown above
Output: 5
Note: There can be more than one mother vertices in a graph. We need to output anyone of them.
For example, in the below graph, vertices 0, 1, and 2 are mother vertices.
Naive Approach: To solve the problem follow the below idea:
A trivial approach will be to perform a DFS/BFS on all the vertices and find whether we can reach all the vertices from that vertex.
Time Complexity: O(V * (E+V))
Auxiliary Space: O(1) It will be O(V) if the stack space for DFS is considered or if we use a BFS.
To solve the problem follow the below idea:
The idea is based on Kosaraju's Strongly Connected Component Algorithm. In a graph of strongly connected components, mother vertices are always vertices of the source components in the component graph. The idea is based on the below fact:
If there exists a mother vertex (or vertices), then one of the mother vertices is the last finished vertex in DFS. (Or a mother vertex has the maximum finish time in DFS traversal). A vertex is said to be finished in DFS if a recursive call for its DFS is over, i.e., all descendants of the vertex have been visited.
How does the above idea work?
Let the last finished vertex be v. Basically, we need to prove that there cannot be an edge from another vertex u to v if u is not another mother vertex (Or there cannot exist a non-mother vertex u such that u-?v is an edge). There can be two possibilities.
- A recursive DFS call is made for u before v. If an edge u-?v exists, then v must have finished before u because v is reachable through u and a vertex finishes after all its descendants.
- A recursive DFS call is made for v before u. In this case also, if an edge u-?v exists, then either v must finish before u (which contradicts our assumption that v is finished at the end) OR u should be reachable from v (which means u is another mother vertex).
Follow the given steps to solve the problem:
- Do DFS traversal of the given graph. While doing traversal keep track of the last finished vertex 'v'. This step takes O(V+E) time.
- If there exists a mother vertex (or vertices), then v must be one (or one of them). Check if v is a mother vertex by doing DFS/BFS from v. This step also takes O(V+E) time.
Below is the implementation of the above approach.
C++
// C++ program to find a mother vertex in O(V+E) time
#include <bits/stdc++.h>
using namespace std;
class Graph {
int V; // No. of vertices
list<int>* adj; // adjacency lists
// A recursive function to print DFS starting from v
void DFSUtil(int v, vector<bool>& visited);
public:
Graph(int V);
void addEdge(int v, int w);
int findMother();
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
// A recursive function to print DFS starting from v
void Graph::DFSUtil(int v, vector<bool>& visited)
{
// Mark the current node as visited and print it
visited[v] = true;
// Recur for all the vertices adjacent to this vertex
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
if (!visited[*i])
DFSUtil(*i, visited);
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
// Returns a mother vertex if exists. Otherwise returns -1
int Graph::findMother()
{
// visited[] is used for DFS. Initially all are
// initialized as not visited
vector<bool> visited(V, false);
// To store last finished vertex (or mother vertex)
int v = 0;
// Do a DFS traversal and find the last finished
// vertex
for (int i = 0; i < V; i++) {
if (visited[i] == false) {
DFSUtil(i, visited);
v = i;
}
}
// If there exist mother vertex (or vertices) in given
// graph, then v must be one (or one of them)
// Now check if v is actually a mother vertex (or graph
// has a mother vertex). We basically check if every
// vertex is reachable from v or not.
// Reset all values in visited[] as false and do
// DFS beginning from v to check if all vertices are
// reachable from it or not.
fill(visited.begin(), visited.end(), false);
DFSUtil(v, visited);
for (int i = 0; i < V; i++)
if (visited[i] == false)
return -1;
return v;
}
// Driver code
int main()
{
// Create a graph given in the above diagram
Graph g(7);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 3);
g.addEdge(4, 1);
g.addEdge(6, 4);
g.addEdge(5, 6);
g.addEdge(5, 2);
g.addEdge(6, 0);
// Function call
cout << "A mother vertex is " << g.findMother();
return 0;
}
// Java program to find a mother
// vertex in O(V+E) time
import java.util.*;
class GFG {
static void addEdge(int u, int v,
ArrayList<ArrayList<Integer> > adj)
{
adj.get(u).add(v);
}
// A recursive function to print DFS starting from v
static void DFSUtil(ArrayList<ArrayList<Integer> > g,
int v, boolean[] visited)
{
// Mark the current node as
// visited and print it
visited[v] = true;
// Recur for all the vertices
// adjacent to this vertex
for (int x : g.get(v)) {
if (!visited[x]) {
DFSUtil(g, x, visited);
}
}
}
// Returns a mother vertex if exists.
// Otherwise returns -1
static int
motherVertex(ArrayList<ArrayList<Integer> > g, int V)
{
// visited[] is used for DFS. Initially
// all are initialized as not visited
boolean[] visited = new boolean[V];
// To store last finished vertex
// (or mother vertex)
int v = -1;
for (int i = 0; i < V; i++) {
if (!visited[i]) {
DFSUtil(g, i, visited);
v = i;
}
}
// If there exist mother vertex (or vertices)
// in given graph, then v must be one
// (or one of them)
// Now check if v is actually a mother
// vertex (or graph has a mother vertex).
// We basically check if every vertex
// is reachable from v or not.
// Reset all values in visited[] as false
// and do DFS beginning from v to check
// if all vertices are reachable from
// it or not.
boolean[] check = new boolean[V];
DFSUtil(g, v, check);
for (boolean val : check) {
if (!val) {
return -1;
}
}
return v;
}
// Driver code
public static void main(String[] args)
{
int V = 7;
int E = 8;
ArrayList<ArrayList<Integer> > adj
= new ArrayList<ArrayList<Integer> >();
for (int i = 0; i < V; i++) {
adj.add(new ArrayList<Integer>());
}
addEdge(0, 1, adj);
addEdge(0, 2, adj);
addEdge(1, 3, adj);
addEdge(4, 1, adj);
addEdge(6, 4, adj);
addEdge(5, 6, adj);
addEdge(5, 2, adj);
addEdge(6, 0, adj);
// Function call
System.out.println("A mother vertex is "
+ motherVertex(adj, V));
}
}
// This code is contributed by Tanay Shah
# Python3 program to find a mother vertex in O(V+E) time
from collections import defaultdict
# This class represents a directed graph using adjacency list
# representation
class Graph:
def __init__(self, vertices):
self.V = vertices # No. of vertices
self.graph = defaultdict(list) # default dictionary
# A recursive function to print DFS starting from v
def DFSUtil(self, v, visited):
# Mark the current node as visited and print it
visited[v] = True
# Recur for all the vertices adjacent to this vertex
for i in self.graph[v]:
if visited[i] == False:
self.DFSUtil(i, visited)
# Add w to the list of v
def addEdge(self, v, w):
self.graph[v].append(w)
# Returns a mother vertex if exists. Otherwise returns -1
def findMother(self):
# visited[] is used for DFS. Initially all are
# initialized as not visited
visited = [False]*(self.V)
# To store last finished vertex (or mother vertex)
v = 0
# Do a DFS traversal and find the last finished
# vertex
for i in range(self.V):
if visited[i] == False:
self.DFSUtil(i, visited)
v = i
# If there exist mother vertex (or vertices) in given
# graph, then v must be one (or one of them)
# Now check if v is actually a mother vertex (or graph
# has a mother vertex). We basically check if every vertex
# is reachable from v or not.
# Reset all values in visited[] as false and do
# DFS beginning from v to check if all vertices are
# reachable from it or not.
visited = [False]*(self.V)
self.DFSUtil(v, visited)
if any(i == False for i in visited):
return -1
else:
return v
# Driver code
if __name__ == '__main__':
g = Graph(7)
g.addEdge(0, 1)
g.addEdge(0, 2)
g.addEdge(1, 3)
g.addEdge(4, 1)
g.addEdge(6, 4)
g.addEdge(5, 6)
g.addEdge(5, 2)
g.addEdge(6, 0)
# Function call
print("A mother vertex is " + str(g.findMother()))
# This code is contributed by Neelam Yadav
// C# program to find a mother
// vertex in O(V+E) time
using System;
using System.Collections.Generic;
class GFG {
static void addEdge(int u, int v, List<List<int> > adj)
{
adj[u].Add(v);
}
// A recursive function to print DFS starting from v
static void DFSUtil(List<List<int> > g, int v,
bool[] visited)
{
// Mark the current node as
// visited and print it
visited[v] = true;
// Recur for all the vertices
// adjacent to this vertex
foreach(int x in g[v])
{
if (!visited[x]) {
DFSUtil(g, x, visited);
}
}
}
// Returns a mother vertex if exists.
// Otherwise returns -1
static int motherVertex(List<List<int> > g, int V)
{
// visited[] is used for DFS. Initially
// all are initialized as not visited
bool[] visited = new bool[V];
// To store last finished vertex
// (or mother vertex)
int v = -1;
for (int i = 0; i < V; i++) {
if (!visited[i]) {
DFSUtil(g, i, visited);
v = i;
}
}
// If there exist mother vertex (or vertices)
// in given graph, then v must be one
// (or one of them)
// Now check if v is actually a mother
// vertex (or graph has a mother vertex).
// We basically check if every vertex
// is reachable from v or not.
// Reset all values in visited[] as false
// and do DFS beginning from v to check
// if all vertices are reachable from
// it or not.
bool[] check = new bool[V];
DFSUtil(g, v, check);
foreach(bool val in check)
{
if (!val) {
return -1;
}
}
return v;
}
// Driver code
public static void Main(String[] args)
{
int V = 7;
// int E = 8;
List<List<int> > adj = new List<List<int> >();
for (int i = 0; i < V; i++) {
adj.Add(new List<int>());
}
addEdge(0, 1, adj);
addEdge(0, 2, adj);
addEdge(1, 3, adj);
addEdge(4, 1, adj);
addEdge(6, 4, adj);
addEdge(5, 6, adj);
addEdge(5, 2, adj);
addEdge(6, 0, adj);
// Function call
Console.WriteLine("A mother vertex is "
+ motherVertex(adj, V));
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript program to find a mother
// vertex in O(V+E) time
function addEdge(u, v, adj)
{
adj[u].push(v);
}
// A recursive function to print DFS starting from v
function DFSUtil(g, v, visited)
{
// Mark the current node as
// visited and print it
visited[v] = true;
// Recur for all the vertices
// adjacent to this vertex
for(let x in g[v])
{
if (!visited[x])
{
DFSUtil(g, x, visited);
}
}
}
// Returns a mother vertex if exists.
// Otherwise returns -1
function motherVertex(g, V)
{
// visited[] is used for DFS. Initially
// all are initialized as not visited
let visited = new Array(V);
for(let i = 0; i < V; i++)
{
visited[i] = false;
}
// To store last finished vertex
// (or mother vertex)
let v = -1;
for(let i = 0; i < V; i++)
{
if (!visited[i])
{
DFSUtil(g, i, visited);
v = i;
}
}
// If there exist mother vertex (or vertices)
// in given graph, then v must be one
// (or one of them)
// Now check if v is actually a mother
// vertex (or graph has a mother vertex).
// We basically check if every vertex
// is reachable from v or not.
// Reset all values in visited[] as false
// and do DFS beginning from v to check
// if all vertices are reachable from
// it or not.
let check = new Array(V);
for(let i = 0; i < V; i++)
{
check[i] = false;
}
DFSUtil(g, v, check);
for(let val in check)
{
if (!val)
{
return -1;
}
}
return v-1;
}
let V = 7;
let E = 8;
let adj = [];
for(let i = 0; i < V; i++)
{
adj.push([]);
}
addEdge(0, 1,adj);
addEdge(0, 2,adj);
addEdge(1, 3,adj);
addEdge(4, 1,adj);
addEdge(6, 4,adj);
addEdge(5, 6,adj);
addEdge(5, 2,adj);
addEdge(6, 0,adj);
document.write("A mother vertex is " + motherVertex(adj, V));
// This code is contributed by divyesh072019.
</script>
OutputA mother vertex is 5
Time Complexity: O(V + E)
Auxiliary Space: O(V)
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