Calculate number of nodes between two vertices in an acyclic Graph by DFS method
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++
// 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 destinationint 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 Graphint 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 Codeint 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
// Java program for the above approachimport java.util.Vector;class GFG{// Function to return the count of nodes// in the path from source to destinationstatic 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 Graphstatic 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 Codepublic 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
# Python3 program for the above approach# Function to return the count of nodes# in the path from source to destinationdef 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 Graphdef 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 Codeif __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#
// C# program for // the above approachusing System;using System.Collections.Generic;class GFG{// Function to return the count of nodes// in the path from source to destinationstatic 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 Graphstatic 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 Codepublic 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 |
Javascript
<script>// Javascript program for the above approach// Function to return the count of nodes// in the path from source to destinationfunction 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 Graphfunction 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 edgeslet V = 8, E = 7;// Given source and destination verticeslet src = 5, dest = 2;// Given edgeslet 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> |
Output:
1
Time Complexity: O(V+E)
Auxiliary Space: O(V)
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