Find the root of the sub-tree whose weighted sum XOR with X is maximum

Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum XOR with given integer X is maximum.

Examples:

Input:

X = 15
Output: 4
Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11
Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) XOR 15 = 7 XOR 15 = 8
Weight of sub-tree for parent 3 = -1 XOR 15 = -16
Weight of sub-tree for parent 4 = 3 XOR 15 = 12
Weight of sub-tree for parent 5 = -2 XOR 15 = -15
Node 4 gives the maximum sub-tree weighted sum XOR X.



Approach: Perform dfs on the tree, and for every node calculate the sub-tree weighted sum rooted at the current node then find the maximum (sum XOR X) value for a node.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
int ans = 0, maxi = INT_MIN;
  
vector<int> graph[100];
vector<int> weight(100);
  
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
void dfs(int node, int parent)
{
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
  
        // Calculating the weighted
        // sum of the subtree
        weight[node] += weight[to];
    }
}
  
// Function to find the node
// having maximum sub-tree sum XOR x
void findMaxX(int n, int x)
{
  
    // For every node
    for (int i = 1; i <= n; i++) {
  
        // If current node's weight XOR x
        // is maximum so far
        if (maxi < (weight[i] ^ x)) {
            maxi = (weight[i] ^ x);
            ans = i;
        }
    }
}
  
// Driver code
int main()
{
    int x = 15;
    int n = 5;
  
    // Weights of the node
    weight[1] = -1;
    weight[2] = 5;
    weight[3] = -1;
    weight[4] = 3;
    weight[5] = -2;
  
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
  
    dfs(1, 1);
    findMaxX(n, x);
  
    cout << ans;
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
    static int ans = 0, maxi = Integer.MIN_VALUE;
  
    static Vector<Integer>[] graph = new Vector[100];
    static Integer[] weight = new Integer[100];
  
    // Function to perform dfs and update the tree
    // such that every node's weight is the sum of
    // the weights of all the nodes in the sub-tree
    // of the current node including itself
    static void dfs(int node, int parent)
    {
        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
  
            // Calculating the weighted
            // sum of the subtree
            weight[node] += weight[to];
        }
    }
  
    // Function to find the node
    // having maximum sub-tree sum XOR x
    static void findMaxX(int n, int x)
    {
  
        // For every node
        for (int i = 1; i <= n; i++)
        {
  
            // If current node's weight XOR x
            // is maximum so far
            if (maxi < (weight[i] ^ x))
            {
                maxi = (weight[i] ^ x);
                ans = i;
            }
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int x = 15;
        int n = 5;
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<Integer>();
          
        // Weights of the node
        weight[1] = -1;
        weight[2] = 5;
        weight[3] = -1;
        weight[4] = 3;
        weight[5] = -2;
  
        // Edges of the tree
        graph[1].add(2);
        graph[2].add(3);
        graph[2].add(4);
        graph[1].add(5);
  
        dfs(1, 1);
        findMaxX(n, x);
  
        System.out.print(ans);
    }
}
  
// This code is contributed by Rajput-Ji

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG
{
    static int ans = 0, maxi = int.MinValue;
  
    static List<int>[] graph = new List<int>[100];
    static int[] weight = new int[100];
  
    // Function to perform dfs and update the tree
    // such that every node's weight is the sum of
    // the weights of all the nodes in the sub-tree
    // of the current node including itself
    static void dfs(int node, int parent)
    {
        foreach (int to in graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
  
            // Calculating the weighted
            // sum of the subtree
            weight[node] += weight[to];
        }
    }
  
    // Function to find the node
    // having maximum sub-tree sum XOR x
    static void findMaxX(int n, int x)
    {
  
        // For every node
        for (int i = 1; i <= n; i++)
        {
  
            // If current node's weight XOR x
            // is maximum so far
            if (maxi < (weight[i] ^ x))
            {
                maxi = (weight[i] ^ x);
                ans = i;
            }
        }
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int x = 15;
        int n = 5;
        for (int i = 0; i < 100; i++)
            graph[i] = new List<int>();
          
        // Weights of the node
        weight[1] = -1;
        weight[2] = 5;
        weight[3] = -1;
        weight[4] = 3;
        weight[5] = -2;
  
        // Edges of the tree
        graph[1].Add(2);
        graph[2].Add(3);
        graph[2].Add(4);
        graph[1].Add(5);
  
        dfs(1, 1);
        findMaxX(n, x);
  
        Console.Write(ans);
    }
}
  
// This code is contributed by PrinciRaj1992

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Output:

4


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Improved By : Rajput-Ji, princiraj1992