查找与X的总和具有最小设置位的节点

给定一棵树，以及所有节点的权重和一个整数x ，任务是找到一个节点i ，以使weight [i] + x给出最小的置位位数，如果两个或多个节点的置位位数相同，则当与x相加，然后找到最小值的那个。
例子：

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 
 

方法：在树上执行dfs并跟踪其与x的总和具有最小设置位的节点。如果两个或更多节点的设置位计数相等，则选择数量最少的一个。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int minimum = INT_MAX, x, ans = INT_MAX;
 
vector graph[100];
vector 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;
}

Java

// Java implementation of the approach
import java.util.*;
class GFG{
 
static int minimum = Integer.MAX_VALUE,
           x, ans = Integer.MAX_VALUE;
 
static Vector []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();
   
  // 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

Python3

# 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

C#

// 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

输出：

1

复杂度分析：

• 时间复杂度： O(N)。
  在dfs中，树的每个节点都处理一次，因此，如果树中总共有N个节点，则由于dfs而导致的复杂度为O(N)。同样，为了处理每个节点，使用了internalin_popcount()函数，该函数的复杂度为O(c)，其中c为常数，并且由于该复杂度为常数，因此不会影响整体时间复杂度。因此，时间复杂度为O(N)。
• 辅助空间： O(1)。
  不需要任何额外的空间，因此空间复杂度是恒定的。

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