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Find the node whose sum with X has maximum set bits

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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:
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>


Output

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:

  1. Initialize a queue to store the nodes for BFS traversal.
  2. Push the root node to the queue.
  3. While the queue is not empty, dequeue a node from the queue.
  4. Calculate the set bits value of the current node and update the maximum and minimum nodes accordingly.
  5. Enqueue all the child nodes of the current node to the queue.
  6. Repeat steps 3 to 5 until the queue is empty.
  7. 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();


Output

4







Complexity Analysis

Time Complexity: O(N) 
Auxiliary Space: O(N)



Last Updated : 16 Oct, 2023
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