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] + x has the maximum set bits. If two or more nodes have the same count of set bits when added with x then find the one with the minimum value.
Examples:
Input:
x = 15
Output: 4
Node 1: setbits(5 + 15) = 2
Node 2: setbits(10 + 15) = 3
Node 3: setbits(11 + 15) = 3
Node 4: setbits(8 + 15) = 4
Node 5: setbits(6 + 15) = 3
Approach: Perform dfs on the tree and keep track of the node whose sum with x has maximum set bits. If two or more nodes have equal count of set bits then choose the one with the minimum number.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
int maximum = INT_MIN, x, ans = INT_MAX;
vector< int > graph[100];
vector< int > weight(100);
// Function to perform dfs to find // the maximum set bits value void dfs( int node, int parent)
{ // If current set bits value is greater than
// the current maximum
int a = __builtin_popcount(weight[node] + x);
if (maximum < a) {
maximum = a;
ans = node;
}
// If count is equal to the maximum
// then choose the node with minimum value
else if (maximum == a)
ans = min(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 = Integer.MAX_VALUE;
static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>();
static Vector<Integer> weight = new Vector<Integer>();
//number of set bits static int __builtin_popcount( int x)
{ int c = 0 ;
for ( int i = 0 ; i < 60 ; i++)
if (((x>>i)& 1 ) != 0 )c++;
return c;
} // Function to perform dfs to find // the maximum value static void dfs( int node, int parent)
{ // If current set bits value is greater than
// the current maximum
int a = __builtin_popcount(weight.get(node) + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
// If count is equal to the maximum
// then choose the node with minimum value
else if (maximum == a)
ans = Math.min(ans, node);
for ( int i = 0 ; i < graph.get(node).size(); i++)
{
if (graph.get(node).get(i) == parent)
continue ;
dfs(graph.get(node).get(i), node);
}
} // Driver code public static void main(String args[])
{ 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 Vector<Integer>());
// Edges of the tree
graph.get( 1 ).add( 2 );
graph.get( 2 ).add( 3 );
graph.get( 2 ).add( 4 );
graph.get( 1 ).add( 5 );
dfs( 1 , 1 );
System.out.println( ans);
} } // This code is contributed by Arnab Kundu |
# Python implementation of the approach from sys import maxsize
maximum, x, ans = - maxsize, None , maxsize
graph = [[] for i in range ( 100 )]
weight = [ 0 ] * 100
# Function to perform dfs to find # the maximum set bits value def dfs(node, parent):
global x, ans, graph, weight, maximum
# If current set bits value is greater than
# the current maximum
a = bin (weight[node] + x).count( '1' )
if maximum < a:
maximum = a
ans = node
# If count is equal to the maximum
# then choose the node with minimum value
elif maximum = = a:
ans = min (ans, 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)
# This code is contributed by # sanjeev2552 |
// 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 >();
// number of set bits static int __builtin_popcount( int x)
{ int c = 0;
for ( int i = 0; i < 60; i++)
if (((x>>i)&1) != 0)c++;
return c;
} // Function to perform dfs to find // the maximum value static void dfs( int node, int parent)
{ // If current set bits value is greater than
// the current maximum
int a = __builtin_popcount(weight[node] + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
// If count is equal to the maximum
// then choose the node with minimum value
else if (maximum == a)
ans = Math.Min(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 mits |
<script> // Javascript implementation of the approach let maximum = Number.MIN_VALUE, x,
ans = Number.MAX_VALUE;
let graph = new Array(100);
for (let i=0;i<100;i++)
{
graph[i]=[];
}
let weight = [];
//number of set bits
function __builtin_popcount(x)
{
let c = 0;
for (let i = 0; i < 60; i++)
if (((x>>i)&1) != 0)
c++;
return c;
}
// Function to perform dfs to find
// the maximum value
function dfs(node,parent)
{
// If current set bits value is greater than
// the current maximum
let a = __builtin_popcount(weight[node] + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
// If count is equal to the maximum
// then choose the node with minimum value
else if (maximum == a)
ans = Math.min(ans, node);
for (let i = 0; i < graph[node].length; i++)
{
if (graph[node][i] == parent)
continue ;
dfs(graph[node][i], node);
}
}
// Driver code
x = 15;
// Weights of the node
weight.push(0);
weight.push(5);
weight.push(10);;
weight.push(11);;
weight.push(8);
weight.push(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 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. Also, for processing each node the builtin_popcount() function is used which has a complexity of O(c) where c is a constant and since this complexity is constant, it does not affect the overall time complexity. Therefore, the time complexity is O(N). -
Auxiliary Space : O(1).
Any extra space is not required, so the space complexity is constant.
Approach 2:
To implement the given problem using BFS approach, we can follow the below steps:
- Initialize a queue to store the nodes for BFS traversal.
- Push the root node to the queue.
- While the queue is not empty, dequeue a node from the queue.
- Calculate the set bits value of the current node and update the maximum and minimum nodes accordingly.
- Enqueue all the child nodes of the current node to the queue.
- Repeat steps 3 to 5 until the queue is empty.
- Print the node with the minimum value.
Here’s the codes for the BFS approach:
#include <bits/stdc++.h> using namespace std;
int maximum = INT_MIN, x, ans = INT_MAX;
vector< int > graph[100];
vector< int > weight(100);
// Function to perform bfs to find // the maximum set bits value void bfs( int root)
{ queue< int > q;
q.push(root);
while (!q.empty()) {
int node = q.front();
q.pop();
// If current set bits value is greater than
// the current maximum
int a = __builtin_popcount(weight[node] + x);
if (maximum < a) {
maximum = a;
ans = node;
}
// If count is equal to the maximum
// then choose the node with minimum value
else if (maximum == a)
ans = min(ans, node);
for ( int to : graph[node]) {
if (to != root) {
q.push(to);
}
}
}
} // 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);
bfs(1);
cout << ans;
return 0;
} |
import java.util.*;
public class Main {
// Initialize variables for maximum, x, and answer
static int maximum = Integer.MIN_VALUE;
static int x;
static int ans = Integer.MAX_VALUE;
// Create an array to represent the graph
static List<Integer>[] graph = new ArrayList[ 100 ];
// Create an array to store node weights
static int [] weight = new int [ 100 ];
// Function to perform BFS to find the maximum set bits value
static void bfs( int root) {
Queue<Integer> q = new LinkedList<>();
q.add(root);
while (!q.isEmpty()) {
int node = q.poll();
// Calculate the set bits value of the current node's weight plus x
int a = Integer.bitCount(weight[node] + x);
// If current set bits value is greater than the current maximum
if (maximum < a) {
maximum = a;
ans = node;
}
// If count is equal to the maximum, choose the node with minimum value
else if (maximum == a) {
ans = Math.min(ans, node);
}
// Check if the neighboring nodes exist
if (graph[node] != null ) {
// Add neighboring nodes to the queue
for ( int to : graph[node]) {
if (to != root) {
q.add(to);
}
}
}
}
}
// Driver code
public static void main(String[] args) {
x = 15 ;
// Weights of the nodes
weight[ 1 ] = 5 ;
weight[ 2 ] = 10 ;
weight[ 3 ] = 11 ;
weight[ 4 ] = 8 ;
weight[ 5 ] = 6 ;
// Initialize the graph edges
graph[ 1 ] = new ArrayList<>(Arrays.asList( 2 , 5 ));
graph[ 2 ] = new ArrayList<>(Arrays.asList( 3 , 4 ));
bfs( 1 );
System.out.println(ans);
}
} |
from queue import Queue
maximum = float ( '-inf' )
x, ans = None , float ( 'inf' )
graph = [[] for _ in range ( 100 )]
weight = [ 0 ] * 100
# Function to perform bfs to find # the maximum set bits value def bfs(root):
global maximum, ans
q = Queue()
q.put(root)
while not q.empty():
node = q.get()
# If current set bits value is greater than
# the current maximum
a = bin (weight[node] + x).count( '1' )
if maximum < a:
maximum = a
ans = node
# If count is equal to the maximum
# then choose the node with minimum value
elif maximum = = a:
ans = min (ans, node)
for to in graph[node]:
if to ! = root:
q.put(to)
# 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 )
bfs( 1 )
print (ans)
|
using System;
using System.Collections.Generic;
public class GFG
{ // Constants to store minimum and maximum values
public static int maximum = int .MinValue;
public static int x, ans = int .MaxValue;
// Create a graph using adjacency list representation
public static List< int >[] graph = new List< int >[100];
// Create a list to store weights of the nodes
public static List< int > weight = new List< int >(100);
// Function to perform BFS to find the maximum set bits value
public static void BFS( int root)
{
Queue< int > q = new Queue< int >();
q.Enqueue(root);
while (q.Count > 0)
{
int node = q.Dequeue();
// If current set bits value is greater than the current maximum
int a = CountSetBits(weight[node] + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
// If count is equal to the maximum, then choose the node with minimum value
else if (maximum == a)
{
ans = Math.Min(ans, node);
}
foreach ( int to in graph[node])
{
if (to != root)
{
q.Enqueue(to);
}
}
}
}
// Function to count the number of set bits in an integer
public static int CountSetBits( int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Driver code
public static void Main()
{
x = 15;
// Weights of the nodes
weight.Add(0); // Dummy value to make it 1-based indexing
weight.Add(5);
weight.Add(10);
weight.Add(11);
weight.Add(8);
weight.Add(6);
// Initialize the graph
for ( int i = 0; i < 100; i++)
{
graph[i] = new List< int >();
}
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
BFS(1);
Console.WriteLine(ans);
}
} |
// Initialize variables for maximum, x, and answer let maximum = Number.MIN_SAFE_INTEGER; let x = 15; let ans = Number.MAX_SAFE_INTEGER; // Create an array to represent the graph let graph = new Array(100);
// Create an array to store node weights let weight = new Array(100);
// Function to perform BFS to find the maximum set bits value function bfs(root) {
let queue = [];
queue.push(root);
while (queue.length > 0) {
let node = queue.shift();
// Calculate the set bits value of the current node's weight plus x
let a = (weight[node] + x).toString(2).split('1').length - 1;
// If current set bits value is greater than the current maximum
if (maximum < a) {
maximum = a;
ans = node;
}
// If count is equal to the maximum, choose the node with minimum value
else if (maximum === a) {
ans = Math.min(ans, node);
}
// Check if the neighboring nodes exist
if (graph[node]) {
// Add neighboring nodes to the queue
for (let i = 0; i < graph[node].length; i++) {
let to = graph[node][i];
if (to !== root) {
queue.push(to);
}
}
}
}
} // Driver code function main() {
// Weights of the nodes
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
// Initialize the graph edges
graph[1] = [2, 5];
graph[2] = [3, 4];
bfs(1);
console.log(ans);
} main(); |
4
Complexity Analysis:
Time Complexity: O(N)
Auxiliary Space: O(N)