Find the node whose sum with X has minimum set bits

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: 1
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 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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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 doesnot affect the overall time comlexity. 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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