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; } |
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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 |
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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 |
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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 |
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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. 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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