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Find the node whose sum with X has minimum set bits
  • Last Updated : 26 Oct, 2020

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 gives the minimum setbits, 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 minimum set bits. If two or more nodes have an equal count of set bits then choose the one with the minimum number.
Below is the implementation of the above approach: 
 



C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int minimum = INT_MAX, x, ans = INT_MAX;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function to perform dfs to find
// the minimum set bits value
void dfs(int node, int parent)
{
    // If current set bits value is smaller than
    // the current minimum
    int a = __builtin_popcount(weight[node] + x);
    if (minimum > a) {
        minimum = a;
        ans = node;
    }
 
    // If count is equal to the minimum
    // then choose the node with minimum value
    else if (minimum == 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;
}

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Java

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// Java implementation of the approach
import java.util.*;
class GFG{
 
static int minimum = Integer.MAX_VALUE,
           x, ans = Integer.MAX_VALUE;
 
static Vector<Integer> []graph =
              new Vector[100];
static int []weight = new int[100];
 
// Function to perform dfs
// to find the minimum set
// bits value
static void dfs(int node,
                int parent)
{
  // If current set bits value
  // is smaller than the current
  // minimum
  int a = Integer.bitCount(weight[node] + x);
  if (minimum > a)
  {
    minimum = a;
    ans = node;
  }
 
  // If count is equal to the
  // minimum then choose the
  // node with minimum value
  else if (minimum == a)
    ans = Math.min(ans, node);
 
  for (int to : graph[node])
  {
    if (to == parent)
      continue;
    dfs(to, node);
  }
}
 
// Driver code
public static void main(String[] args)
{
  x = 15;
  for (int i = 0; i < graph.length; i++)
    graph[i] = new Vector<Integer>();
   
  // 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].add(2);
  graph[2].add(3);
  graph[2].add(4);
  graph[1].add(5);
 
  dfs(1, 1);
 
  System.out.print(ans);
}
}
 
// This code is contributed by gauravrajput1

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Python3

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# Python3 implementation of the approach
from sys import maxsize
 
minimum, x, ans = maxsize, None, maxsize
 
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function to perform dfs to find
# the minimum set bits value
def dfs(node, parent):
    global x, ans, graph, weight, minimum
 
    # If current set bits value is greater than
    # the current minimum
    a = bin(weight[node] + x).count('1')
 
    if minimum > a:
        minimum = a
        ans = node
 
    # If count is equal to the minimum
    # then choose the node with minimum value
    elif minimum == 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

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C#

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// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
 
class GFG{
     
static int minimum = int.MaxValue, x,
               ans = int.MaxValue;
 
static ArrayList[] graph = new ArrayList[100];
static int[] weight = new int[100];
 
static int PopCount(int n)
{
    int count = 0;
     
    while (n > 0)
    {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Function to perform dfs to find
// the minimum set bits value
static void dfs(int node, int parent)
{
     
    // If current set bits value is smaller
    // than the current minimum
    int a = PopCount(weight[node] + x);
    if (minimum > a)
    {
        minimum = a;
        ans = node;
    }
 
    // If count is equal to the minimum
    // then choose the node with minimum value
    else if (minimum == a)
        ans = Math.Min(ans, node);
 
    foreach(int to in graph[node])
    {
        if (to == parent)
            continue;
             
        dfs(to, node);
    }
}
     
// Driver Code
public static void Main(string[] args)
{
    x = 15;
     
    for(int i = 0; i < 100; i++)
        graph[i] = new ArrayList();
     
    // 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].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 rutvik_56

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Output: 

1


 

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 doesn’t 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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