Encuentre el Node cuyo xor con x da el valor máximo

Dado un árbol, y los pesos de todos los Nodes y un número entero x , la tarea es encontrar un Node i tal que el peso[i] x o x sea máximo.
Ejemplos: 
 

Aporte: 
 

x = 15 
Salida: 1 
Node 1: 5 xor 15 = 10 
Node 2: 10 xor 15 = 5 
Node 3: 11 xor 15 = 4 
Node 4: 8 xor 15 = 7 
Node 5: 6 xor 15 = 9 
 

Enfoque: Realice dfs en el árbol y realice un seguimiento del Node cuyo xor ponderado con x da el valor máximo.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int maximum = INT_MIN, x, ans;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function to perform dfs to find
// the maximum xored value
void dfs(int node, int parent)
{
    // If current value is less than
    // the current maximum
    if (maximum < (weight[node] ^ x)) {
        maximum = weight[node] ^ x;
        ans = node;
    }
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    x = 15;
 
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
 
    // 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);
 
    cout << ans;
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    static int maximum = Integer.MIN_VALUE, x, ans;
 
    @SuppressWarnings("unchecked")
    static Vector<Integer>[] graph = new Vector[100];
    static int[] weight = new int[100];
 
    // This block is executed even before main() function
    // This is necessary otherwise this program will
    // throw "NullPointerException"
    static
    {
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<>();
    }
 
    // Function to perform dfs to find
    // the maximum xored value
    static void dfs(int node, int parent)
    {
 
        // If current value is less than
        // the current maximum
        if (maximum < (weight[node] ^ x))
        {
            maximum = weight[node] ^ x;
            ans = node;
        }
        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        x = 15;
 
        // Weights of the node
        weight[1] = 5;
        weight[2] = 10;
        weight[3] = 11;
        weight[4] = 8;
        weight[5] = 6;
 
        // Edges of the tree
        graph[1].add(2);
        graph[2].add(3);
        graph[2].add(4);
        graph[1].add(5);
 
        dfs(1, 1);
 
        System.out.println(ans);
    }
}
 
// This code is contributed by
// sanjeev2552

# Python3 implementation of the approach
import sys
maximum = -sys.maxsize - 1
graph = [[0 for i in range(100)]
            for j in range(100)]
weight = [0 for i in range(100)]
ans = []
 
# Function to perform dfs to find
# the maximum xored value
def dfs(node, parent):
    global maximum
     
    # If current value is less than
    # the current maximum
    if (maximum < (weight[node] ^ x)):
        maximum = weight[node] ^ x
        ans.append(node)
         
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
         
# Driver code
if __name__ == '__main__':
    x = 15
 
    # Weights of the node
    weight[1] = 5
    weight[2] = 10
    weight[3] = 11
    weight[4] = 8
    weight[5] = 6
 
    # Edges of the tree
    graph[1].append(2)
    graph[2].append(3)
    graph[2].append(4)
    graph[1].append(5)
 
    dfs(1, 1)
 
    print(ans[0])
     
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
static int maximum = int.MinValue, x,
ans = int.MaxValue;
 
static List<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();
 
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current value is less than
    // the current maximum
    if (maximum < (weight[node] ^ x))
    {
        maximum = weight[node] ^ x;
        ans = node;
    }
         
    for (int i = 0; i < graph[node].Count; i++)
    {
        if (graph[node][i] == parent)
            continue;
        dfs(graph[node][i], node);
    }
}
 
// Driver code
public static void Main()
{
    x = 15;
 
    // Weights of the node
    weight.Add(0);
    weight.Add(5);
    weight.Add(10);
    weight.Add(11);;
    weight.Add(8);
    weight.Add(6);
     
    for(int i = 0; i < 100; i++)
    graph.Add(new List<int>());
 
    // Edges of the tree
    graph[1].Add(2);
    graph[2].Add(3);
    graph[2].Add(4);
    graph[1].Add(5);
 
    dfs(1, 1);
    Console.Write( ans);
}
}
 
// This code is contributed by SHUBHAMSINGH10

<script>
// Javascript implementation of the approach
 
 
let maximum = Number.MIN_SAFE_INTEGER;
let ans = [];
 
let graph = new Array();
 
for(let i = 0; i < 100; i++){
    graph.push(new Array().fill(0));
}
 
let weight = new Array(100).fill(0);
 
 
// Function to perform dfs to find
// the maximum xored value
function dfs(node, parent) {
    // If current value is less than
    // the current maximum
    if (maximum < (weight[node] ^ x)) {
        maximum = weight[node] ^ x;
        ans = node;
    }
    for (let to of graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
 
let x = 15;
 
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
 
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
 
dfs(1, 1);
 
document.write(ans);
 
// This code is contributed by gfgking
</script>

1

Análisis de Complejidad: 
 

• Complejidad temporal: O(N). 
  En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Por lo tanto, la complejidad del tiempo es O(N).
• Espacio Auxiliar : O(1). 
  No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.