Find the node whose xor with x gives maximum value
Given a tree, and the weights of all the nodes and an integer x, the task is to find a node i such that weight[i] xor x is maximum.
Examples:
Input:
x = 15
Output: 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
Approach: Perform dfs on the tree and keep track of the node whose weighted xor with x gives the maximum value.
Below is the implementation of the above approach:
C++
// 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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
1
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(1).
Any extra space is not required, so the space complexity is constant.
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