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 valuevoid 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 codeint 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 approachimport 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 approachfrom sys import maxsize# Function to perform dfs to find# the minimum valuedef 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 Codeif __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.
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