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++
// 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
// 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 |
Python3
# 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#
// 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 |
Javascript
<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:
C++
#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; } |
Java
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); } } |
Python3
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) |
C#
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); } } |
Javascript
// 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)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!