Count the nodes of the tree whose weighted string contains a vowel

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights contain a vowel.

Examples:

Input:

Output: 2
Only the strings of the nodes 1 and 5 contain vowels.



Approach: Perform dfs on the tree and for every node, check if it’s string contains vowels, If yes then increment the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
int cnt = 0;
  
vector<int> graph[100];
vector<string> weight(100);
  
// Function that returns true
// if the string contains any vowel
bool containsVowel(string str)
{
    for (int i = 0; i < str.length(); i++) {
        char ch = tolower(str[i]);
        if (ch == 'a' || ch == 'e' || ch == 'i'
            || ch == 'o' || ch == 'u')
            return true;
    }
    return false;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
  
    // Weight of the current node
    string x = weight[node];
  
    // If the weight contains any vowel
    if (containsVowel(x))
        cnt += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
  
    // Weights of the node
    weight[1] = "geek";
    weight[2] = "btch";
    weight[3] = "bcb";
    weight[4] = "by";
    weight[5] = "mon";
  
    // 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 << cnt;
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.*;
  
class GFG
{
  
    static int cnt = 0
      
    static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>(); 
    static Vector<String> weight = new Vector<String>(); 
      
    // Function that returns true 
    // if the String contains any vowel 
    static boolean containsVowel(String str) 
    
        for (int i = 0; i < str.length(); i++) 
        
            char ch = str.charAt(i);
            if(ch < 97)ch += 32;
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u'
                return true
        
        return false
    
      
    // Function to perform dfs 
    static void dfs(int node, int parent) 
    
      
        // Weight of the current node 
        String x = weight.get(node); 
      
        // If the weight contains any vowel 
        if (containsVowel(x)) 
            cnt += 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[])
    
      
        // Weights of the node 
        weight.add(""); 
        weight.add("geek"); 
        weight.add( "btch"); 
        weight.add( "bcb"); 
        weight.add( "by"); 
        weight.add( "mon"); 
          
        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( cnt); 
      
    
}
  
// This code is contributed by andrew1234

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Python

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# Python3 implementation of the approach
cnt = 0
  
graph = [[] for i in range(100)]
weight = [0 for i in range(100)]
  
# Function that returns True
# if the contains any vowel
def containsVowel(Str):
  
    for i in range(len(Str)):
        ch = Str[i]
        if (ch == 'a' or ch == 'e' or ch == 'i' or 
                            ch == 'o' or ch == 'u'):
            return True
  
    return False
  
  
# Function to perform dfs
def dfs(node, parent):
    global cnt
  
    # Weight of the current node
    x = weight[node]
  
    # If the weight contains any vowel
    if (containsVowel(x)):
        cnt += 1
  
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code
  
# Weights of the node
weight[1] = "geek"
weight[2] = "btch"
weight[3] = "bcb"
weight[4] = "by"
weight[5] = "mon"
  
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
  
dfs(1, 1)
  
print(cnt)
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the approach 
using System;
using System.Collections.Generic; 
  
class GFG
{
  
    static int cnt = 0; 
      
    static List<List<int>> graph = new List<List<int>>(); 
    static List<String> weight = new List<String>(); 
      
    // Function that returns true 
    // if the String contains any vowel 
    static Boolean containsVowel(String str) 
    
        for (int i = 0; i < str.Length; i++) 
        
            char ch = str[i];
            if(ch < 97)
                ch += (char)32;
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u'
                return true
        
        return false
    
      
    // Function to perform dfs 
    static void dfs(int node, int parent) 
    
      
        // Weight of the current node 
        String x = weight[node]; 
      
        // If the weight contains any vowel 
        if (containsVowel(x)) 
            cnt += 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)
    
      
        // Weights of the node 
        weight.Add(""); 
        weight.Add("geek"); 
        weight.Add( "btch"); 
        weight.Add( "bcb"); 
        weight.Add( "by"); 
        weight.Add( "mon"); 
          
        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( cnt); 
      
    
}
  
// This code has been contributed by 29AjayKumar

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Output:

2


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