Given a tree, and the weights of all the nodes, the task is to count the nodes whose weights are divisible by x.
Examples:
Input:
x = 5
Output: 2
Only the nodes 1 and 2 have weights divisible by 5.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by x or not. If yes then increment the count.
Implementation:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;long ans = 0;int x;vector<int> graph[100];vector<int> weight(100);// Function to perform dfsvoid dfs(int node, int parent){ // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); }}// Driver codeint main(){ x = 5; // 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 long ans = 0; static int x; static Vector<Vector<Integer>> graph=new Vector<Vector<Integer>>(); static Vector<Integer> weight=new Vector<Integer>(); // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node // is divisible by x if (weight.get(node) % x == 0) ans += 1; 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 = 5; // 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
# Python3 implementation of the approach ans = 0graph = [[] for i in range(100)]weight = [0] * 100# Function to perform dfs def dfs(node, parent): global ans,x # If weight of the current node # is divisible by x if (weight[node] % x == 0): ans += 1 for to in graph[node]: if (to == parent): continue dfs(to, node)# Driver code x = 5# Weights of the node weight[1] = 5weight[2] = 10weight[3] = 11weight[4] = 8weight[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 SHUBHAMSINGH10 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG{ static long ans = 0; static int x; static List<List<int>> graph = new List<List<int>>(); static List<int> weight = new List<int>(); // Function to perform dfs static void dfs(int node, int parent) { // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; 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 = 5; // 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 contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach let ans = 0;let x;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 dfsfunction dfs(node, parent){ // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; for(let to=0;to<graph[node].length;to++) { if(graph[node][to] == parent) continue dfs(graph[node][to], node); }}// Driver code x = 5; // 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 Dharanendra L V. </script> |
Output:
2
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) when 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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