Count nodes having Bitwise XOR of all edges in their path from the root equal to K

Last Updated : 18 Aug, 2021

Given a Binary Tree consisting of N nodes and two integers R and K. Each edge of the tree has a positive integer associated with it, given in the form {u, v, w} where the edge (u, v) has weight w. The task is to calculate the number of nodes S having Bitwise XOR of all edges in the path from root R to S is equal to K.

Examples:

Input: R = 1, K = 0, N = 7, Edges[][] = {{1, 2, 3}, {1, 3, 1}, {2, 4, 3}, {2, 5, 4}, {3, 6, 1}, {3, 7, 2}}
Output: 2
Explanation:
Representation of the given Binary Tree:

The following pair of nodes have a Bitwise XOR of edges in the path connecting them as K = 0:
Pair 1: (1, 4) = (3 ^ 3) = 0
Pair 2: (1, 6) = (1 ^ 1) = 0

Input: R = 1, K = 0, N = 9, Edges[][] = {{1, 2, 3}, {1, 3, 2}, {2, 4, 3}, {2, 5, 4}, {3, 6, 1}, {3, 7, 2}, {6, 8, 3}, {6, 9, 7}}
Output: 3
Explanation:
The representation of given Binary Tree is as follows:

The following pair of nodes have a Bitwise XOR of edges in the path connecting them as K = 0:
Pair 1: (1, 4) = (3 ^ 3) = 0
Pair 2: (1, 8) = (2 ^ 1 ^ 3) = 0
Pair 3: (1, 7) = (2 ^ 2) = 0

Approach: The problem can be solved using the Depth First Search approach. Follow the steps below to solve the problem:

Initialize the variable ans and xor with 0 to store the number of pairs and the current xor of edges.
Traverse the given tree using Depth First Search starting from the given root vertex R.
For every node u, visit its adjacent nodes.
For each edge {u, v}, if xor is equal to K, increment ans by 1. Otherwise, for the current edge {u, v, w}, update xor as xor = (xor^w) where ^ is the bitwise XOR.
After traversing, print the value stored in the counter ans as the number of pairs.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Initialize the adjacency list
// to represent the tree
vector<pair<int, int> > adj[100005];

// Marks visited / unvisited vertices
int visited[100005] = { 0 };

// Stores the required count of nodes
int ans = 0;

// DFS to visit each vertex
void dfs(int node, int xorr, int k)
{
    // Mark the current node
    // as visited
    visited[node] = 1;

    // Update the counter xor is K
    if (node != 1 && xorr == k)
        ans++;

    // Visit adjacent nodes
    for (auto x : adj[node]) {

        if (!visited[x.first]) {

            // Calculate Bitwise XOR of
            // edges in the path
            int xorr1 = xorr ^ x.second;

            // Recursive call to dfs function
            dfs(x.first, xorr1, k);
        }
    }
}

// Function to construct the tree and
// print required count of nodes
void countNodes(int N, int K, int R,
                vector<vector<int> > edges)
{

    // Add edges
    for (int i = 0; i < N - 1; i++) {
        int u = edges[i][0], v = edges[i][1],
            w = edges[i][2];
        adj[u].push_back({ v, w });
        adj[v].push_back({ u, w });
    }

    dfs(R, 0, K);

    // Print answer
    cout << ans << "\n";
}

// Driver Code
int main()
{
    // Given K and R
    int K = 0, R = 1;

    // Given edges
    vector<vector<int> > edges
        = { { 1, 2, 3 }, { 1, 3, 1 }, 
            { 2, 4, 3 }, { 2, 5, 4 }, 
            { 3, 6, 1 }, { 3, 7, 2 } };

    // Number of vertices
    int N = edges.size();

    // Function call
    countNodes(N, K, R, edges);

    return 0;
}

Java

// Java program for the 
// above approach
import java.util.*;
class GFG{

static class pair
{ 
  int first, second; 
  public pair(int first, 
              int second)  
  { 
    this.first = first; 
    this.second = second; 
  }    
} 
  
// Initialize the adjacency list
// to represent the tree
static Vector<pair> []adj = 
       new Vector[100005];

// Marks visited / unvisited
// vertices
static int visited[] = 
       new int[100005];

// Stores the required 
// count of nodes
static int ans = 0;

// DFS to visit each 
// vertex
static void dfs(int node, 
                int xorr, 
                int k)
{
  // Mark the current node
  // as visited
  visited[node] = 1;

  // Update the counter 
  // xor is K
  if (node != 1 && 
      xorr == k)
    ans++;

  // Visit adjacent nodes
  for (pair x : adj[node]) 
  {
    if (visited[x.first] != 1) 
    {
      // Calculate Bitwise XOR of
      // edges in the path
      int xorr1 = xorr ^ x.second;

      // Recursive call to dfs
      // function
      dfs(x.first, xorr1, k);
    }
  }
}

// Function to construct the tree and
// print required count of nodes
static void countNodes(int N, int K,
                       int R, int[][] edges)
{
  // Add edges
  for (int i = 0; i < N - 1; i++)
  {
    int u = edges[i][0], 
        v = edges[i][1],
    w = edges[i][2];
    adj[u].add(new pair(v, w ));
    adj[v].add(new pair(u, w ));
  }

  dfs(R, 0, K);

  // Print answer
  System.out.print(ans + "\n");
}

// Driver Code
public static void main(String[] args)
{
  // Given K and R
  int K = 0, R = 1;
  
  for (int i = 0; i < adj.length; i++)
    adj[i] = new Vector<pair>();
  // Given edges
  int[][] edges = {{1, 2, 3}, 
                   {1, 3, 1},  
                   {2, 4, 3}, 
                   {2, 5, 4}, 
                   {3, 6, 1}, 
                   {3, 7, 2}};

  // Number of vertices
  int N = edges.length;

  // Function call
  countNodes(N, K, R, edges);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Initialize the adjacency list
# to represent the tree
adj = [[] for i in range(100005)]

# Marks visited / unvisited vertices
visited = [0] * 100005

# Stores the required count of nodes
ans = 0

# DFS to visit each vertex
def dfs(node, xorr, k):
    
    global ans
    
    # Mark the current node
    # as visited
    visited[node] = 1

    # Update the counter xor is K
    if (node != 1 and xorr == k):
        ans += 1

    # Visit adjacent nodes
    for x in adj[node]:
        if (not visited[x[0]]):

            # Calculate Bitwise XOR of
            # edges in the path
            xorr1 = xorr ^ x[1]

            # Recursive call to dfs function
            dfs(x[0], xorr1, k)

# Function to construct the tree and
# prrequired count of nodes
def countNodes(N, K, R, edges):

    # Add edges
    for i in range(N - 1):
        u = edges[i][0]
        v = edges[i][1]
        w = edges[i][2]
        
        adj[u].append([v, w])
        adj[v].append([u, w])

    dfs(R, 0, K)

    # Print answer
    print(ans)

# Driver Code
if __name__ == '__main__':
    
    # Given K and R
    K = 0
    R = 1

    # Given edges
    edges = [ [ 1, 2, 3 ],[ 1, 3, 1 ],
              [ 2, 4, 3 ],[ 2, 5, 4 ],
              [ 3, 6, 1 ],[ 3, 7, 2 ] ]

    # Number of vertices
    N = len(edges)

    # Function call
    countNodes(N, K, R, edges)

# This code is contributed by mohit kumar 29

// C# program for the 
// above approach
using System;
using System.Collections.Generic;
class GFG{

public class pair
{ 
  public int first, 
             second; 
  public pair(int first, 
              int second)  
  { 
    this.first = first; 
    this.second = second; 
  }    
} 
  
// Initialize the adjacency list
// to represent the tree
static List<pair> []adj = 
       new List<pair>[100005];

// Marks visited / unvisited
// vertices
static int []visited = 
       new int[100005];

// Stores the required 
// count of nodes
static int ans = 0;

// DFS to visit each 
// vertex
static void dfs(int node, 
                int xorr, 
                int k)
{
  // Mark the current node
  // as visited
  visited[node] = 1;

  // Update the counter 
  // xor is K
  if (node != 1 && 
      xorr == k)
    ans++;

  // Visit adjacent nodes
  foreach (pair x in adj[node]) 
  {
    if (visited[x.first] != 1) 
    {
      // Calculate Bitwise XOR of
      // edges in the path
      int xorr1 = xorr ^ x.second;

      // Recursive call to dfs
      // function
      dfs(x.first, xorr1, k);
    }
  }
}

// Function to construct the tree and
// print required count of nodes
static void countNodes(int N, int K,
                       int R, int[,] edges)
{
  // Add edges
  for (int i = 0; i < N - 1; i++)
  {
    int u = edges[i,0];
     int   v = edges[i,1],
    w = edges[i,2];
    adj[u].Add(new pair(v, w ));
    adj[v].Add(new pair(u, w ));
  }

  dfs(R, 0, K);

  // Print answer
  Console.Write(ans + "\n");
}

// Driver Code
public static void Main(String[] args)
{
  // Given K and R
  int K = 0, R = 1;
  
  for (int i = 0; i < adj.Length; i++)
    adj[i] = new List<pair>();
  
  // Given edges
  int[,] edges = {{1, 2, 3}, 
                   {1, 3, 1},  
                   {2, 4, 3}, 
                   {2, 5, 4}, 
                   {3, 6, 1}, 
                   {3, 7, 2}};

  // Number of vertices
  int N = edges.GetLength(0);

  // Function call
  countNodes(N, K, R, edges);
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>
// Javascript program for the above approach

// Initialize the adjacency list
// to represent the tree
let adj = [];
for (let i = 0; i < 100005; i++) {
    adj.push([])
}

// Marks visited / unvisited vertices
let visited = new Array(100005).fill(0);

// Stores the required count of nodes
let ans = 0;

// DFS to visit each vertex
function dfs(node, xorr, k) {
    // Mark the current node
    // as visited
    visited[node] = 1;

    // Update the counter xor is K
    if (node != 1 && xorr == k)
        ans++;

    // Visit adjacent nodes
    for (let x of adj[node]) {

        if (!visited[x[0]]) {

            // Calculate Bitwise XOR of
            // edges in the path
            let xorr1 = xorr ^ x[1];

            // Recursive call to dfs function
            dfs(x[0], xorr1, k);
        }
    }
}

// Function to construct the tree and
// print required count of nodes
function countNodes(N, K, R, edges) {

    // Add edges
    for (let i = 0; i < N - 1; i++) {
        let u = edges[i][0], v = edges[i][1],
            w = edges[i][2];
        adj[u].push([v, w]);
        adj[v].push([u, w]);
    }

    dfs(R, 0, K);

    // Print answer
    document.write(ans + "<br>");
}

// Driver Code

// Given K and R
let K = 0, R = 1;

// Given edges
let edges
    = [[1, 2, 3], [1, 3, 1],
       [2, 4, 3], [2, 5, 4],
       [3, 6, 1], [3, 7, 2]];

// Number of vertices
let N = edges.length;

// Function call
countNodes(N, K, R, edges);




// This code is contributed by _saurabh_jaiswal
</script>

Output:

Time Complexity: O(N) where N is the number of nodes.
Auxiliary Space: O(N)

Count pair of nodes with greater Bitwise AND than Bitwise XOR in given Linked List

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Count nodes having Bitwise XOR of all edges in their path from the root equal to K

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