Count number of times each Edge appears in all possible paths of a given Tree
Given an Undirected Connected Graph in the form of a tree consisting of N nodes and (N – 1) edges, the task for each edge is to count the number of times it appears across all possible paths in the Tree.
Examples:
Input:
Output: 3 4 3
Explanation:
All possible paths of a given tree are {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}
Edge 1 occurs in the paths {(1, 2), (1, 3), (1, 4)}. Therefore, the frequency of the edge is 3.
Edge 2 occurs in the paths {(1, 3), (1, 4), (2, 3), (2, 4)}. Therefore, the frequency of the edge is 4.
Edge 3 occurs in the paths {(1, 4), (2, 4), (3, 4)}. Therefore, the frequency of the edge is 3.
Input:
Output: 4 6 4 4
Explanation:
Edge 1 occurs in the paths {(1, 2), (1, 3), (1, 4), (1, 5)}. Therefore, the frequency of the edge is 4
Edge 2 occurs in the paths {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}. Therefore, the frequency of the edge is 6
Edge 3 occurs in the paths {(1, 4), (2, 4), (3, 4), (4, 5)}. Therefore, the frequency of the edge is 4
Edge 4 occurs in the paths {(1, 5), (2, 5), (3, 5), (4, 5)}. Therefore, the frequency of the edge is 4
Naive Approach: The simplest approach is to generate all possible paths from each node of the given graph and store the count of edges occurring in these paths by a HashMap. Finally, print the frequencies of each edge.
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the following observation needs to be made:
The green-colored edge will appear in all the paths that connect any vertex from the subtree on its left to any vertex from the subtree on its right.
Therefore, the number of paths in which the edge occurs = Product of the count of nodes in the two subtrees = 5 * 3 = 15.
Follow the steps below in order to solve the problem:
- Root the tree at any random vertex, say 1.
- Perform DFS at Root. Using DFS calculate the subtree size connected to the edges.
- The frequency of each edge connected to subtree is (subtree size) * (N – subtree size).
- Store the value calculated above for each node in a HashMap. Finally, after complete the traversal of the tree, traverse the HashMap to print the result.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Number of nodes int N; // Structure of a Node struct Node { int node; int edgeLabel; }; // Adjacency List to // represent the Tree vector<Node> adj[100005]; // Stores the frequencies // of every edge vector< int > freq; // Function to perform DFS int dfs( int u = 1, int p = 1) { // Add the current node to // size of subtree rooted at u int sz = 1; // Iterate over its children for ( auto a : adj[u]) { // Check if child is not parent if (a.node != p) { // Get the subtree size // for the child int val = dfs(a.node, u); // Set the frequency // of the current edge freq[a.edgeLabel] = val * (N - val); // Add the subtree size // to itself sz += val; } } // Return the subtree size return sz; } // Function to add edge between nodes void addEdge( int u, int v, int label) { adj[u].push_back({ v, label }); adj[v].push_back({ u, label }); } // Function to print the frequencies // of each edge in all possible paths void printFrequencies() { // Stores the frequency // of all the edges freq = vector< int >(N); // Perform DFS dfs(); for ( int i = 1; i < N; i++) { cout << freq[i] << " " ; } } // Driver Code int main() { N = 4; addEdge(1, 2, 1); addEdge(2, 3, 2); addEdge(3, 4, 3); printFrequencies(); return 0; } |
Java
// Java Program to implement // the above approach import java.util.*; class GFG{ // Number of nodes static int N; // Structure of a Node static class Node { int node; int edgeLabel; public Node( int node, int edgeLabel) { super (); this .node = node; this .edgeLabel = edgeLabel; } }; // Adjacency List to // represent the Tree static Vector<Node> []adj = new Vector[ 100005 ]; // Stores the frequencies // of every edge static int []freq; // Function to perform DFS static int dfs( int u , int p) { // Add the current node to // size of subtree rooted at u int sz = 1 ; // Iterate over its children for (Node a : adj[u]) { // Check if child is not parent if (a.node != p) { // Get the subtree size // for the child int val = dfs(a.node, u); // Set the frequency // of the current edge freq[a.edgeLabel] = val * (N - val); // Add the subtree size // to itself sz += val; } } // Return the subtree size return sz; } // Function to add edge between nodes static void addEdge( int u, int v, int label) { adj[u].add( new Node( v, label )); adj[v].add( new Node( u, label)); } // Function to print the frequencies // of each edge in all possible paths static void printFrequencies() { // Stores the frequency // of all the edges freq = new int [N]; // Perform DFS dfs( 1 , 1 ); for ( int i = 1 ; i < N; i++) { System.out.print(freq[i] + " " ); } } // Driver Code public static void main(String[] args) { N = 4 ; for ( int i = 0 ; i < adj.length; i++) adj[i] = new Vector<Node>(); addEdge( 1 , 2 , 1 ); addEdge( 2 , 3 , 2 ); addEdge( 3 , 4 , 3 ); printFrequencies(); } } // This code is contributed by shikhasingrajput |
Python3
# Python3 program to implement # the above approach # Number of nodes N = 4 # Structure of a Node class Node: def __init__( self , v, label): self .node = v self .edgeLabel = label # Adjacency list to # represent the Tree adj = [] for i in range ( 100005 ): adj.append([]) # Stores the frequencies # of each edge freq = [ 0 ] * N # Function to perform DFS def dfs(u = 1 , p = 1 ): global N # Add the current node to # size of subtree rooted at u sz = 1 # Iterate over its children for a in adj[u]: # Check if child is not parent if a.node ! = p: # Get the subtree size # for the child val = dfs(a.node, u) # Set the frequency # of the current edge freq[a.edgeLabel] = val * (N - val) # Add the subtree size # to itself sz + = val # Return the subtree size return sz # Function to add edge between nodes def addEdge(u, v, label): adj[u].append(Node(v, label)) adj[v].append(Node(u, label)) # Function to print the frequencies # of each edge in all possible paths def printFrequencies(): # Stores the frequency # of all the edges global N # Perform DFS dfs() for i in range ( 1 , N): print (freq[i], end = " " ) # Driver code N = 4 addEdge( 1 , 2 , 1 ) addEdge( 2 , 3 , 2 ) addEdge( 3 , 4 , 3 ) printFrequencies() # This code is contributed by Stuti Pathak |
C#
// C# Program to implement // the above approach using System; using System.Collections.Generic; class GFG{ // Number of nodes static int N; // Structure of a Node public class Node { public int node; public int edgeLabel; public Node( int node, int edgeLabel) { this .node = node; this .edgeLabel = edgeLabel; } }; // Adjacency List to // represent the Tree static List<Node> []adj = new List<Node>[100005]; // Stores the frequencies // of every edge static int []freq; // Function to perform DFS static int dfs( int u, int p) { // Add the current node to // size of subtree rooted at u int sz = 1; // Iterate over its children foreach (Node a in adj[u]) { // Check if child is not parent if (a.node != p) { // Get the subtree size // for the child int val = dfs(a.node, u); // Set the frequency // of the current edge freq[a.edgeLabel] = val * (N - val); // Add the subtree size // to itself sz += val; } } // Return the subtree size return sz; } // Function to add edge between nodes static void addEdge( int u, int v, int label) { adj[u].Add( new Node(v, label)); adj[v].Add( new Node(u, label)); } // Function to print the frequencies // of each edge in all possible paths static void printFrequencies() { // Stores the frequency // of all the edges freq = new int [N]; // Perform DFS dfs(1, 1); for ( int i = 1; i < N; i++) { Console.Write(freq[i] + " " ); } } // Driver Code public static void Main(String[] args) { N = 4; for ( int i = 0; i < adj.Length; i++) adj[i] = new List<Node>(); addEdge(1, 2, 1); addEdge(2, 3, 2); addEdge(3, 4, 3); printFrequencies(); } } // This code is contributed by gauravrajput1 |
Javascript
<script> // Javascript Program to implement the above approach // Number of nodes let N; // Structure of a Node class Node { constructor(node, edgeLabel) { this .node = node; this .edgeLabel = edgeLabel; } } // Adjacency List to // represent the Tree let adj = new Array(100005); // Stores the frequencies // of every edge let freq; // Function to perform DFS function dfs(u, p) { // Add the current node to // size of subtree rooted at u let sz = 1; // Iterate over its children for (let a = 0; a < adj[u].length; a++) { // Check if child is not parent if (adj[u][a].node != p) { // Get the subtree size // for the child let val = dfs(adj[u][a].node, u); // Set the frequency // of the current edge freq[adj[u][a].edgeLabel] = val * (N - val); // Add the subtree size // to itself sz += val; } } // Return the subtree size return sz; } // Function to add edge between nodes function addEdge(u, v, label) { adj[u].push( new Node( v, label )); adj[v].push( new Node( u, label)); } // Function to print the frequencies // of each edge in all possible paths function printFrequencies() { // Stores the frequency // of all the edges freq = new Array(N); // Perform DFS dfs(1, 1); for (let i = 1; i < N; i++) { document.write(freq[i] + " " ); } } N = 4; for (let i = 0; i < adj.length; i++) adj[i] = []; addEdge(1, 2, 1); addEdge(2, 3, 2); addEdge(3, 4, 3); printFrequencies(); </script> |
3 4 3
Time Complexity: O(N)
Auxiliary Space: O(N)