Find the node whose absolute difference with X gives minimum value

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| is minimum.
Examples:

Input:

x = 15 
Output: 3 
Node 1: |5 – 15| = 10 
Node 2: |10 – 15| = 5 
Node 3: |11 -15| = 4 
Node 4: |8 – 15| = 7 
Node 5: |6 -15| = 9 

Approach: Perform dfs on the tree and keep track of the node whose weighted absolute difference with x gives the minimum value.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int minimum = INT_MAX, x, ans;

vector<int> graph[100];
vector<int> weight(100);

// Function to perform dfs to find
// the minimum value
void dfs(int node, int parent)
{
    // If current value is less than
    // the current minimum
    if (minimum > abs(weight[node] - x)) {
        minimum = abs(weight[node] - x);
        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.*;
import java.lang.*;

class GFG
{

    static int minimum = Integer.MAX_VALUE, x, ans;

    @SuppressWarnings("unchecked")
    static Vector<Integer>[] graph = new Vector[100];
    static int[] weight = new int[100];

    // This block is executed even before main() function
    // This is necessary otherwise this program will
    // throw "NullPointerException"
    static
    {
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<>();
    }

    // Function to perform dfs to find
    // the minimum xored value
    static void dfs(int node, int parent)
    {

        // If current value is less than
        // the current minimum
        if (minimum > Math.abs(weight[node] - x))
        {
            minimum = Math.abs(weight[node] - x);
            ans = node;
        }
        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }

    // Driver Code
    public static void main(String[] args)
    {
        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].add(2);
        graph[2].add(3);
        graph[2].add(4);
        graph[1].add(5);

        dfs(1, 1);

        System.out.println(ans);
    }
}

// This code is contributed by SHUBHAMSINGH10

Python3

# Python3 implementation of the approach
from sys import maxsize

# Function to perform dfs to find
# the minimum value
def dfs(node, parent):
    global minimum, graph, weight, x, ans

    # If current value is less than
    # the current minimum
    if minimum > abs(weight[node] - x):
        minimum = abs(weight[node] - x)
        ans = node

    for to in graph[node]:
        if to == parent:
            continue
        dfs(to, node)

# Driver Code
if __name__ == "__main__":
    minimum = maxsize
    graph = [[] for i in range(100)]
    weight = [0] * 100
    x = 15
    ans = 0

    # 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 minimum = int.MaxValue, x, ans;

static List<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();

// Function to perform dfs to find
// the minimum value
static void dfs(int node, int parent)
{
    // If current value is more than
    // the current minimum
    if (minimum > Math.Abs(weight[node] - x))
    {
        minimum = Math.Abs(weight[node] - x);
        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(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 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.WriteLine( ans);
}
}

// This code is contributed by shubhamsingh10

Javascript

<script>

// Javascript implementation of the approach

    let minimum = Number.MAX_VALUE, x, ans;

    let graph = new Array(100);


    let weight = new Array(100);
    for(let i=0;i<100;i++)
    {
        graph[i]=[];
        weight[i]=0;
    }

    // Function to perform dfs to find
    // the minimum xored value
    function dfs(node,parent)
    {
        // If current value is less than
        // the current minimum
        if (minimum > Math.abs(weight[node] - x))
        {
            minimum = Math.abs(weight[node] - x);
            ans = node;
        }
        for(let to=0;to<graph[node].length;to++)
        {
            if(graph[node][to] == parent)
                continue
            dfs(graph[node][to], node);
        }

    }

    // Driver code

    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(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:

3

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. Therefore, the time complexity is O(N).
• Auxiliary Space : O(1). 
  Any extra space is not required, so the space complexity is constant.