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Find count of pair of nodes at even distance | Set 2 (Using BFS)

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  • Last Updated : 28 Jun, 2022

Given a connected acyclic graph with N nodes numbered from 1 to N and N-1 edges, find out the pair of nodes that are at even distance from each other.

Note: The graph is represented in the form of an adjacency list.

Examples:

Input: N = 3, graph = {{}, {2}, {1, 3}, {2}}
Output:1
Explanation: Here there are three pairs {1, 2}, {1, 3}
and {2, 3} and only {1, 3} has even distance between them.
i.e.,         1
            /
         2
      /
   3

Input: N = 5, graph = {{}, {2, 4}, {1, 3}, {2}, {1, 5}, {4}}
Output: 4
Explanation: There are four pairs {1, 3}, {1, 5}, {2, 4}
and {3, 5} which has even distance.

 

Approach: The DFS approach of this article is already discussed in Set-1 of this article. Here we will be using the BFS Traversal Of Graph to solve the problem based on the following idea:

Start traversing from any node (say 1) as the root of the graph and store the number of nodes in odd and even numbered level. The nodes in even numbered level are distance away from each other. The same is true for nodes in odd numbered levels.

Follow the steps mentioned below to implement the idea:

  • Create an array to keep track of all the visited nodes.
  • Using queue do BFS traversal of the graph.
  • Initially push the first node in the queue and then traverse and push its neighbours which are not yet visited into the queue and continue the BFS in this way.
  • Count the number of nodes in even numbered levels (say X) and odd numbered levels (say Y).
  • The answer will be the sum of the number of ways to choose any two elements from X (i.e (X * (X – 1)) / 2)and any two elements from Y (i.e. (Y * (Y – 1)) / 2 ).

Below is the implementation of the above approach:

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the count of nodes at
// even distance using BFS method
int countOfNodes(vector<int> graph[], int n)
{
    // Declare one vector to check that
    // element is visited or not
    vector<int> vis(n + 1);
 
    // Declare one queue
    queue<pair<int, int> > q;
    long long _0 = 0, _1 = 0;
 
    // Initially push the first node with its
    // level as even in the queue
    q.push({ 1, 0 });
 
    // Run this loop until q is not empty
    while (!q.empty()) {
        vis[q.front().first] = 1;
 
        // Check for the adjacent nodes
        for (auto child : graph[q.front().first]) {
            // Check only if adjacent node
            // is not visited
            if (!vis[child])
                q.push({ child, 1 - q.front().second });
        }
        if (q.front().second)
            _1++;
        else
            _0++;
        q.pop();
    }
 
    // Answer will be the sum of the number
    // of ways to choose any two elements
    // of even and odd type
    int ans
        = (_0 * (_0 - 1)) / 2
          + (_1 * (_1 - 1)) / 2;
    return ans;
}
 
// Driver code
int main()
{
    int N = 3;
 
    // Creating adjacency list for the graph
    vector<int> graph[N + 1];
    graph[1].push_back(2);
    graph[2].push_back(1);
    graph[2].push_back(3);
    graph[3].push_back(2);
 
    // Function call
    cout << countOfNodes(graph, N);
    return 0;
}

Java




// Java code to implement the approach
import java.io.*;
import java.util.*;
 
// User defined Pair class
class Pair {
  int x;
  int y;
 
  // Constructor
  public Pair(int x, int y)
  {
    this.x = x;
    this.y = y;
  }
}
 
class GFG {
 
  static void addEdge(ArrayList<ArrayList<Integer> > adj,
                      int u, int v)
  {
    adj.get(u).add(v);
  }
 
  // Function to find the count of nodes at
  // even distance using BFS method
  public static long
    countOfNodes(ArrayList<ArrayList<Integer> > graph,
                 int n)
  {
    // Declare one vector to check that
    // element is visited or not
    int vis[] = new int[n + 1];
 
    // Declare one queue
 
    Queue<Pair> q = new LinkedList<>();
 
    long _0 = 0, _1 = 0;
 
    // Initially push the first node with its
    // level as even in the queue
    q.add(new Pair(1, 0));
 
    // Run this loop until q is not empty
    while (!q.isEmpty()) {
      vis[q.peek().x] = 1;
 
      // Check for the adjacent nodes
      for (Integer child : graph.get(q.peek().x)) {
        // Check only if adjacent node
        // is not visited
        if (vis[child] == 0)
          q.add(new Pair(child, 1 - q.peek().y));
      }
      if (q.peek().y != 0)
        _1++;
      else
        _0++;
      q.poll();
    }
 
    // Answer will be the sum of the number
    // of ways to choose any two elements
    // of even and odd type
    long ans
      = (_0 * (_0 - 1)) / 2 + (_1 * (_1 - 1)) / 2;
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 3;
 
    // Creating adjacency list for the graph
    ArrayList<ArrayList<Integer> > graph
      = new ArrayList<ArrayList<Integer> >(N + 1);
 
    for (int i = 0; i < N + 1; i++)
      graph.add(new ArrayList<Integer>());
    addEdge(graph, 1, 2);
    addEdge(graph, 2, 1);
    addEdge(graph, 2, 3);
    addEdge(graph, 3, 2);
 
    // Function call
    System.out.print(countOfNodes(graph, N));
  }
}
 
// This code is contributed by Rohit Pradhan

Python3




# Python3 code to implement the above approach
 
# Function to find the count of nodes at
# even distance using BFS method
def countOfNodes(graph, n):
 
    # Declare one vector to check that
    # element is visited or not
    vis = [None] * (n + 1)
 
    # Declare one queue
    q = []
    _0 = 0
    _1 = 0
 
    # Initially push the first node with its
    # level as even in the queue
    q.append([1, 0])
 
    # Run this loop until q is not empty
    while (len(q) > 0):
        front = q.pop(0)
        vis[front[0]] = 1
 
        # Check for the adjacent nodes
        for child in graph[front[0]]:
            # Check only if adjacent node
            # is not visited
            if vis[child] is None:
                q.append([child, 1 - front[1]])
 
        if (front[1]):
            _1 += 1
        else:
            _0 += 1
 
    # Answer will be the sum of the number
    # of ways to choose any two elements
    # of even and odd type
    ans = (_0 * (_0 - 1)) // 2 + (_1 * (_1 - 1)) // 2
    return ans
 
# Driver code
N = 3
 
# Creating adjacency list for the graph
graph = [[]]
graph.append([2])
graph.append([1, 3])
graph.append([2])
 
# Function call
print(countOfNodes(graph, N))
 
# This code is contributed by phasing17

C#




// C# code to implement the approach
using System;
using System.Collections.Generic;
 
// User defined Pair class
class Pair {
  public int x;
  public int y;
 
  // Constructor
  public Pair(int x, int y)
  {
    this.x = x;
    this.y = y;
  }
}
 
class GFG {
 
  static void addEdge(List<List<int> > adj, int u, int v)
  {
    adj[u].Add(v);
  }
 
  // Function to find the count of nodes at
  // even distance using BFS method
  public static long countOfNodes(List<List<int> > graph,
                                  int n)
  {
    // Declare one vector to check that
    // element is visited or not
    int[] vis = new int[n + 1];
 
    // Declare one queue
 
    Queue<Pair> q = new Queue<Pair>();
 
    long _0 = 0, _1 = 0;
 
    // Initially push the first node with its
    // level as even in the queue
    q.Enqueue(new Pair(1, 0));
 
    // Run this loop until q is not empty
    while (q.Count > 0) {
      vis[q.Peek().x] = 1;
 
      // Check for the adjacent nodes
      foreach(int child in graph[q.Peek().x])
      {
        // Check only if adjacent node
        // is not visited
        if (vis[child] == 0)
          q.Enqueue(
          new Pair(child, 1 - q.Peek().y));
      }
      if (q.Peek().y != 0)
        _1++;
      else
        _0++;
      q.Dequeue();
    }
 
    // Answer will be the sum of the number
    // of ways to choose any two elements
    // of even and odd type
    long ans
      = (_0 * (_0 - 1)) / 2 + (_1 * (_1 - 1)) / 2;
    return ans;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int N = 3;
 
    // Creating adjacency list for the graph
    List<List<int> > graph
      = new List<List<int> >(N + 1);
 
    for (int i = 0; i < N + 1; i++)
      graph.Add(new List<int>());
    addEdge(graph, 1, 2);
    addEdge(graph, 2, 1);
    addEdge(graph, 2, 3);
    addEdge(graph, 3, 2);
 
    // Function call
    Console.Write(countOfNodes(graph, N));
  }
}
 
// This code is contributed by phasing17

Javascript




<script>
 
// JavaScript code to implement the above approach
 
// Function to find the count of nodes at
// even distance using BFS method
function countOfNodes(graph, n)
{
 
    // Declare one vector to check that
    // element is visited or not
    let vis = new Array(n + 1);
 
    // Declare one queue
    let q = [];
    let _0 = 0, _1 = 0;
 
    // Initially push the first node with its
    // level as even in the queue
    q.push([1, 0 ]);
 
    // Run this loop until q is not empty
    while (q.length>0) {
        let front = q.shift();
        vis[front[0]] = 1;
 
        // Check for the adjacent nodes
        for (let child of graph[front[0]]) {
            // Check only if adjacent node
            // is not visited
            if (!vis[child])
                q.push([ child, 1 - front[1] ]);
        }
        if (front[1])
            _1++;
        else
            _0++;
    }
 
    // Answer will be the sum of the number
    // of ways to choose any two elements
    // of even and odd type
    let ans
        = (_0 * (_0 - 1)) / 2
        + (_1 * (_1 - 1)) / 2;
    return ans;
}
 
// Driver code
let N = 3;
 
// Creating adjacency list for the graph
let graph = new Array(N + 1).fill([]).map(()=>new Array());
graph[1].push(2);
graph[2].push(1);
graph[2].push(3);
graph[3].push(2);
 
// Function call
document.write(countOfNodes(graph, N));
 
// This code is contributed by shinjanpatra
 
</script>

Output

1

Time Complexity: O(V+E) = O(N). As V = number of nodes = N, E = number of edges = N-1 as given.
Auxiliary Space: O(N)


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