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++ 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;
} |
// 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 |
# Python implementation of the approach from 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 itself def 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 x def 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 code ans = 0
maxi = - maxsize
graph = [[] for i in range ( 100 )]
weight = [ 0 ] * 100
x = 15
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 ].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# 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 |
<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> |
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.