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Find the node whose sum with X has maximum set bits
  • Last Updated : 20 Apr, 2021

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.

 

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