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 minimum.
Examples:
Input:
X = 15
Output: 5
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 3 gives the minimum 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 minimum (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, mini = INT_MAX;
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 minimum sub-tree sum XOR x
void findMinX(int n, int x)
{
// For every node
for (int i = 1; i <= n; i++) {
// If current node's weight XOR x
// is minimum so far
if (mini > (weight[i] ^ x)) {
mini = (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);
findMinX(n, x);
cout << ans;
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
static int ans = 0, mini = Integer.MAX_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 minimum sub-tree sum XOR x
static void findMinX(int n, int x)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight XOR x
// is minimum so far
if (mini > (weight[i] ^ x))
{
mini = (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);
findMinX(n, x);
System.out.print(ans);
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation of the approach
ans = 0
mini = 2**32
graph = [[] for i in range(100)]
weight = [0] * 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
def dfs(node, parent):
global ans, mini, graph, weight, x
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 minimum sub-tree sum XOR x
def findMinX(n, x):
global ans, mini,graph,weight
# For every node
for i in range(1, n + 1):
# If current node's weight XOR x
# is minimum so far
if (mini > (weight[i] ^ x)):
mini = (weight[i] ^ x)
ans = i
# Driver code
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)
findMinX(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, mini = int.MaxValue;
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 minimum sub-tree sum XOR x
static void findMinX(int n, int x)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight XOR x
// is minimum so far
if (mini > (weight[i] ^ x))
{
mini = (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);
findMinX(n, x);
Console.Write(ans);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript implementation of the approach
let ans = 0, mini = Number.MAX_VALUE;
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=0;to<graph[node].length;to++)
{
if (graph[node][to] == parent)
continue;
dfs(graph[node][to], node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum XOR x
function findMinX(n,X)
{
// For every node
for (let i = 1; i <= n; i++)
{
// If current node's weight XOR x
// is minimum so far
if (mini > (weight[i] ^ x))
{
mini = (weight[i] ^ x);
ans = i;
}
}
}
// Driver code
let 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);
findMinX(n, x);
document.write(ans);
// This code is contributed by unknown2108
</script>
Output:
3
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 space can be up to O(n).