Find the node whose absolute difference with X gives maximum 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 maximum.
Examples:
Input:
x = 15
Output: 1
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 maximum value.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; int maximum = INT_MIN, x, ans; vector< int > graph[100]; vector< int > weight(100); // Function to perform dfs to find // the maximum value void dfs( int node, int parent) { // If current value is more than // the current maximum if (maximum < abs (weight[node] - x)) { maximum = 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.*; class GFG { static int maximum = Integer.MIN_VALUE, x, ans; static Vector<Vector<Integer>> graph= new Vector<Vector<Integer>>(); static Vector<Integer> weight= new Vector<Integer>(); // Function to perform dfs to find // the maximum value static void dfs( int node, int parent) { // If current value is more than // the current maximum if (maximum < Math.abs(weight.get(node) - x)) { maximum = Math.abs(weight.get(node) - x); ans = node; } for ( int i = 0 ; i < graph.get(node).size(); i++) { if (graph.get(node).get(i) == parent) continue ; dfs(graph.get(node).get(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 Vector<Integer>()); // Edges of the tree graph.get( 1 ).add( 2 ); graph.get( 2 ).add( 3 ); graph.get( 2 ).add( 4 ); graph.get( 1 ).add( 5 ); dfs( 1 , 1 ); System.out.println( ans); } } // This code is contributed by Arnab Kundu |
Python3
# Python 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 maximum = int .MinValue, 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 maximum value static void dfs( int node, int parent) { // If current value is more than // the current maximum if (maximum < Math.Abs(weight[node] - x)) { maximum = 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 Princi Singh |
Javascript
<script> // Javascript implementation of the approach let maximum = Number.MIN_VALUE, x, ans; let graph= []; let weight=[]; // Function to perform dfs to find // the maximum value function dfs(node,parent) { // If current value is more than // the current maximum if (maximum < Math.abs(weight[node] - x)) { maximum = Math.abs(weight[node] - x); ans = node; } for (let i = 0; i < graph[node].length; i++) { if (graph[node][i] == parent) continue ; dfs(graph[node][i], node); } } // Driver code x = 15; // Weights of the node weight.push(0); weight.push(5); weight.push(10);; weight.push(11);; weight.push(8); weight.push(6); for (let i = 0; i < 100; i++) graph.push([]); // 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:
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. 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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