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

• Last Updated : 04 Jun, 2021

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:

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Input:

Output:
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++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `int` `cnt = 0;` `vector<``int``> graph[100];``vector 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);` `    ``// Function call``    ``dfs(1, 1);` `    ``cout << cnt;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG {` `    ``static` `int` `cnt = ``0``;` `    ``static` `Vector > graph``        ``= ``new` `Vector >();``    ``static` `Vector weight = ``new` `Vector();` `    ``// 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());` `        ``// Edges of the tree``        ``graph.get(``1``).add(``2``);``        ``graph.get(``2``).add(``3``);``        ``graph.get(``2``).add(``4``);``        ``graph.get(``1``).add(``5``);` `        ``// Function call``        ``dfs(``1``, ``1``);` `        ``System.out.println(cnt);``    ``}``}` `// This code is contributed by andrew1234`

## Python3

 `# 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``)` `# Function call``dfs(``1``, ``1``)` `print``(cnt)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``static` `int` `cnt = 0;` `    ``static` `List > graph = ``new` `List >();``    ``static` `List weight = ``new` `List();` `    ``// 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);` `        ``// Function call``        ``dfs(1, 1);` `        ``Console.WriteLine(cnt);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 ``
Output
`2`

Complexity Analysis:

Time Complexity: O(N*Len) where Len is the maximum length of the weighted string of a node in the given tree.
In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) for N nodes in the tree. Also, processing of every node involves traversing the weighted string of that node once, thus adding a complexity of O(Len) where Len is the length of the weighted string. Therefore, the total time complexity is O(N*Len).

Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.

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