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Count number of times each Edge appears in all possible paths of a given Tree

  • Difficulty Level : Hard
  • Last Updated : 07 Sep, 2021
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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>
Output: 
3 4 3

 

Time Complexity: O(N) 
Auxiliary Space: O(N) 

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