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++
// 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 itselfvoid 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 xvoid 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 codeint 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;} |
Java
// Java implementation of the approachimport 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 |
Python
# Python implementation of the approachfrom sys import maxsize# 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 itselfdef dfs(node, parent): global maxi, graph, weight, x, ans for 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 xdef findMaxX(n, x): global maxi, graph, weight, ans # For every node for i in range(1, n + 1): # If current node's weight XOR x # is maximum so far if (maxi < (weight[i] ^ x)): maxi = (weight[i] ^ x) ans = i# Driver codeans = 0maxi = -maxsizegraph = [[] for i in range(100)]weight = [0]*100x = 15n = 5# Weights of the nodeweight[1] = -1weight[2] = 5weight[3] = -1weight[4] = 3weight[5] = -2# Edges of the treegraph[1].append(2)graph[2].append(3)graph[2].append(4)graph[1].append(5)dfs(1, 1)findMaxX(n, x)print(ans)# This code is contributed by SHUBHAMSINGH10 |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// Javascript implementation of the approach let maxi = Number.MIN_VALUE, x, ans; let graph = new Array(100); let weight = new Array(100); for(let i = 0; i < 100; i++) { graph[i] = []; weight[i] = 0; } // 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 function dfs(node, parent) { for(let 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 function findMaxX(n, x) { // For every node for (let 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 x = 15; let 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(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); dfs(1, 1); findMaxX(n, x); document.write(ans); // This code is contributed by unknown2108</script> |
Output:
4
Complexity Analysis:
- Time Complexity : O(N).
In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Therefore, the time complexity is O(N). - Auxiliary Space : O(n).
Recursion Stack.
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