Given a connected acyclic graph consisting of V vertices and E edges, a source vertex src, and a destination vertex dest, the task is to count the number of vertices between the given source and destination vertex in the graph.
Examples:
Input: V = 8, E = 7, src = 7, dest = 8, edges[][] ={{1 4}, {4, 5}, {4, 2}, {2, 6}, {6, 3}, {2, 7}, {3, 8}}
Output: 3
Explanation:
The path between 7 and 8 is 7 -> 2 -> 6 -> 3 -> 8.
So, the number of nodes between 7 and 8 is 3.Input: V = 8, E = 7, src = 5, dest = 2, edges[][] ={{1 4}, {4, 5}, {4, 2}, {2, 6}, {6, 3}, {2, 7}, {3, 8}}
Output: 1
Explanation:
The path between 5 and 2 is 5 -> 4 -> 2.
So, the number of nodes between 5 and 2 is 1.
Approach: The problem can also be solved using the Disjoint Union method as stated in this article. Another approach to this problem is to solve using the Depth First Search method. Follow the steps below to solve this problem:
- Initialize a visited array vis[] to mark which nodes are already visited. Mark all the nodes as 0, i.e., not visited.
- Perform a DFS to find the number of nodes present in the path between src and dest.
- The number of nodes between src and dest is equal to the difference between the length of the path between them and 2, i.e., (pathSrcToDest – 2).
- Since the graph is acyclic and connected, there will always be a single path between src and dest.
Below is the implementation of the above algorithm.
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to return the count of nodes // in the path from source to destination int dfs( int src, int dest, int * vis,
vector< int >* adj)
{ // Mark the node visited
vis[src] = 1;
// If dest is reached
if (src == dest) {
return 1;
}
// Traverse all adjacent nodes
for ( int u : adj[src]) {
// If not already visited
if (!vis[u]) {
int temp = dfs(u, dest, vis, adj);
// If there is path, then
// include the current node
if (temp != 0) {
return temp + 1;
}
}
}
// Return 0 if there is no path
// between src and dest through
// the current node
return 0;
} // Function to return the // count of nodes between two // given vertices of the acyclic Graph int countNodes( int V, int E, int src, int dest,
int edges[][2])
{ // Initialize an adjacency list
vector< int > adj[V + 1];
// Populate the edges in the list
for ( int i = 0; i < E; i++) {
adj[edges[i][0]].push_back(edges[i][1]);
adj[edges[i][1]].push_back(edges[i][0]);
}
// Mark all the nodes as not visited
int vis[V + 1] = { 0 };
// Count nodes in the path from src to dest
int count = dfs(src, dest, vis, adj);
// Return the nodes between src and dest
return count - 2;
} // Driver Code int main()
{ // Given number of vertices and edges
int V = 8, E = 7;
// Given source and destination vertices
int src = 5, dest = 2;
// Given edges
int edges[][2]
= { { 1, 4 }, { 4, 5 },
{ 4, 2 }, { 2, 6 },
{ 6, 3 }, { 2, 7 },
{ 3, 8 } };
cout << countNodes(V, E, src, dest, edges);
return 0;
} |
// Java program for the above approach import java.util.Vector;
class GFG{
// Function to return the count of nodes // in the path from source to destination static int dfs( int src, int dest, int []vis,
Vector<Integer> []adj)
{ // Mark the node visited
vis[src] = 1 ;
// If dest is reached
if (src == dest)
{
return 1 ;
}
// Traverse all adjacent nodes
for ( int u : adj[src])
{
// If not already visited
if (vis[u] == 0 )
{
int temp = dfs(u, dest,
vis, adj);
// If there is path, then
// include the current node
if (temp != 0 )
{
return temp + 1 ;
}
}
}
// Return 0 if there is no path
// between src and dest through
// the current node
return 0 ;
} // Function to return the // count of nodes between two // given vertices of the acyclic Graph static int countNodes( int V, int E,
int src, int dest,
int edges[][])
{ // Initialize an adjacency list
Vector<Integer> []adj = new Vector[V + 1 ];
for ( int i = 0 ; i < adj.length; i++)
adj[i] = new Vector<Integer>();
// Populate the edges in the list
for ( int i = 0 ; i < E; i++)
{
adj[edges[i][ 0 ]].add(edges[i][ 1 ]);
adj[edges[i][ 1 ]].add(edges[i][ 0 ]);
}
// Mark all the nodes as
// not visited
int vis[] = new int [V + 1 ];
// Count nodes in the path
// from src to dest
int count = dfs(src, dest,
vis, adj);
// Return the nodes
// between src and dest
return count - 2 ;
} // Driver Code public static void main(String[] args)
{ // Given number of vertices and edges
int V = 8 , E = 7 ;
// Given source and destination vertices
int src = 5 , dest = 2 ;
// Given edges
int edges[][] = {{ 1 , 4 }, { 4 , 5 },
{ 4 , 2 }, { 2 , 6 },
{ 6 , 3 }, { 2 , 7 },
{ 3 , 8 }};
System.out.print(countNodes(V, E,
src, dest,
edges));
} } // This code is contributed by shikhasingrajput |
# Python3 program for the above approach # Function to return the count of nodes # in the path from source to destination def dfs(src, dest, vis, adj):
# Mark the node visited
vis[src] = 1
# If dest is reached
if (src = = dest):
return 1
# Traverse all adjacent nodes
for u in adj[src]:
# If not already visited
if not vis[u]:
temp = dfs(u, dest, vis, adj)
# If there is path, then
# include the current node
if (temp ! = 0 ):
return temp + 1
# Return 0 if there is no path
# between src and dest through
# the current node
return 0
# Function to return the # count of nodes between two # given vertices of the acyclic Graph def countNodes(V, E, src, dest, edges):
# Initialize an adjacency list
adj = [[] for i in range (V + 1 )]
# Populate the edges in the list
for i in range (E):
adj[edges[i][ 0 ]].append(edges[i][ 1 ])
adj[edges[i][ 1 ]].append(edges[i][ 0 ])
# Mark all the nodes as not visited
vis = [ 0 ] * (V + 1 )
# Count nodes in the path from src to dest
count = dfs(src, dest, vis, adj)
# Return the nodes between src and dest
return count - 2
# Driver Code if __name__ = = '__main__' :
# Given number of vertices and edges
V = 8
E = 7
# Given source and destination vertices
src = 5
dest = 2
# Given edges
edges = [ [ 1 , 4 ], [ 4 , 5 ],
[ 4 , 2 ], [ 2 , 6 ],
[ 6 , 3 ], [ 2 , 7 ],
[ 3 , 8 ] ]
print (countNodes(V, E, src, dest, edges))
# This code is contributed by mohit kumar 29 |
// C# program for // the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to return the count of nodes // in the path from source to destination static int dfs( int src, int dest,
int []vis, List< int > []adj)
{ // Mark the node visited
vis[src] = 1;
// If dest is reached
if (src == dest)
{
return 1;
}
// Traverse all adjacent nodes
foreach ( int u in adj[src])
{
// If not already visited
if (vis[u] == 0)
{
int temp = dfs(u, dest,
vis, adj);
// If there is path, then
// include the current node
if (temp != 0)
{
return temp + 1;
}
}
}
// Return 0 if there is no path
// between src and dest through
// the current node
return 0;
} // Function to return the // count of nodes between two // given vertices of the acyclic Graph static int countNodes( int V, int E,
int src, int dest,
int [,]edges)
{ // Initialize an adjacency list
List< int > []adj = new List< int >[V + 1];
for ( int i = 0; i < adj.Length; i++)
adj[i] = new List< int >();
// Populate the edges in the list
for ( int i = 0; i < E; i++)
{
adj[edges[i, 0]].Add(edges[i, 1]);
adj[edges[i, 1]].Add(edges[i, 0]);
}
// Mark all the nodes as
// not visited
int []vis = new int [V + 1];
// Count nodes in the path
// from src to dest
int count = dfs(src, dest,
vis, adj);
// Return the nodes
// between src and dest
return count - 2;
} // Driver Code public static void Main(String[] args)
{ // Given number of vertices and edges
int V = 8, E = 7;
// Given source and destination vertices
int src = 5, dest = 2;
// Given edges
int [,]edges = {{1, 4}, {4, 5},
{4, 2}, {2, 6},
{6, 3}, {2, 7},
{3, 8}};
Console.Write(countNodes(V, E, src,
dest, edges));
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program for the above approach // Function to return the count of nodes // in the path from source to destination function dfs(src,dest,vis,adj)
{ // Mark the node visited
vis[src] = 1;
// If dest is reached
if (src == dest)
{
return 1;
}
// Traverse all adjacent nodes
for (let u=0;u< adj[src].length;u++)
{
// If not already visited
if (vis[adj[src][u]] == 0)
{
let temp = dfs(adj[src][u], dest,
vis, adj);
// If there is path, then
// include the current node
if (temp != 0)
{
return temp + 1;
}
}
}
// Return 0 if there is no path
// between src and dest through
// the current node
return 0;
} // Function to return the // count of nodes between two // given vertices of the acyclic Graph function countNodes(V,E,src,dest,edges)
{ // Initialize an adjacency list
let adj = new Array(V + 1);
for (let i = 0; i < adj.length; i++)
adj[i] = [];
// Populate the edges in the list
for (let i = 0; i < E; i++)
{
adj[edges[i][0]].push(edges[i][1]);
adj[edges[i][1]].push(edges[i][0]);
}
// Mark all the nodes as
// not visited
let vis = new Array(V + 1);
for (let i=0;i<vis.length;i++)
{
vis[i]=0;
}
// Count nodes in the path
// from src to dest
let count = dfs(src, dest,
vis, adj);
// Return the nodes
// between src and dest
return count - 2;
} // Driver Code // Given number of vertices and edges let V = 8, E = 7; // Given source and destination vertices let src = 5, dest = 2; // Given edges let edges = [[1, 4], [4, 5], [4, 2], [2, 6], [6, 3], [2, 7], [3, 8]]; document.write(countNodes(V, E, src, dest,
edges));
// This code is contributed by unknown2108 </script> |
1
Time Complexity: O(V+E)
Auxiliary Space: O(V)