Count the nodes of a tree whose weighted string is an anagram of the given string

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weighted string is an anagram with the given string str.

Examples:

Input:

str = “geek”
Output: 2
Only the weighted strings of the nodes 2 and 6
are anagrams of the given string “geek”.

Approach: Perform dfs on the tree and for every node, check if it’s weighted string is anagram with the given string or not, If not 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;
  
string s;
int cnt = 0;
  
vector<int> graph[100];
vector<string> weight(100);
  
// Function that return true if both
// the strings are anagram of each other
bool anagram(string x, string s)
{
    sort(x.begin(), x.end());
    sort(s.begin(), s.end());
    if (x == s)
        return true;
    else
        return false;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If current node's weighted
    // string is an anagram of
    // the given string s
    if (anagram(weight[node], s))
        cnt += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    s = "geek";
  
    // Weights of the nodes
    weight[1] = "eeggk";
    weight[2] = "geek";
    weight[3] = "gekrt";
    weight[4] = "tree";
    weight[5] = "eetr";
    weight[6] = "egek";
  
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
    graph[5].push_back(6);
  
    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 String s;
    static int cnt = 0;
  
    static Vector<Integer>[] graph = new Vector[100];
    static String[] weight = new String[100];
  
    // Function that return true if both
    // the Strings are anagram of each other
    static boolean anagram(String x, String s) 
    {
        x = sort(x);
        s = sort(s);
        if (x.equals(s))
            return true;
        else
            return false;
    }
  
    static String sort(String inputString) 
    {
        // convert input string to char array
        char tempArray[] = inputString.toCharArray();
  
        // sort tempArray
        Arrays.sort(tempArray);
  
        // return new sorted string
        return new String(tempArray);
    }
  
    // Function to perform dfs
    static void dfs(int node, int parent)
    {
        // If current node's weighted
        // String is an anagram of
        // the given String s
        if (anagram(weight[node], s))
            cnt += 1;
  
        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        s = "geek";
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<Integer>();
          
        // Weights of the nodes
        weight[1] = "eeggk";
        weight[2] = "geek";
        weight[3] = "gekrt";
        weight[4] = "tree";
        weight[5] = "eetr";
        weight[6] = "egek";
  
        // Edges of the tree
        graph[1].add(2);
        graph[2].add(3);
        graph[2].add(4);
        graph[1].add(5);
        graph[5].add(6);
  
        dfs(1, 1);
  
        System.out.print(cnt);
    }
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
cnt = 0
  
graph = [[] for i in range(100)]
weight = [0] * 100
  
# Function that return true if both 
# the strings are anagram of each other 
def anagram(x, s):
    x = sorted(list(x)) 
    s = sorted(list(s))
    if (x == s):
        return True
    else:
        return False
  
# Function to perform dfs 
def dfs(node, parent):
    global cnt, s
      
    # If weight of the current node 
    # string is an anagram of 
    # the given string s 
    if (anagram(weight[node], s)):
        cnt += 1
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code 
s = "geek"
  
# Weights of the nodes 
weight[1] = "eeggk"
weight[2] = "geek"
weight[3] = "gekrt"
weight[4] = "tree"
weight[5] = "eetr"
weight[6] = "egek"
  
# Edges of the tree 
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
  
dfs(1, 1)
print(cnt)
  
# This code is contributed by SHUBHAMSINGH10

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    static String s;
    static int cnt = 0;
  
    static List<int>[] graph = new List<int>[100];
    static String[] weight = new String[100];
  
    // Function that return true if both
    // the Strings are anagram of each other
    static bool anagram(String x, String s) 
    {
        x = sort(x);
        s = sort(s);
        if (x.Equals(s))
            return true;
        else
            return false;
    }
  
    static String sort(String inputString) 
    {
        // convert input string to char array
        char []tempArray = inputString.ToCharArray();
  
        // sort tempArray
        Array.Sort(tempArray);
  
        // return new sorted string
        return new String(tempArray);
    }
  
    // Function to perform dfs
    static void dfs(int node, int parent)
    {
        // If current node's weighted
        // String is an anagram of
        // the given String s
        if (anagram(weight[node], s))
            cnt += 1;
  
        foreach (int to in graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        s = "geek";
        for (int i = 0; i < 100; i++)
            graph[i] = new List<int>();
          
        // Weights of the nodes
        weight[1] = "eeggk";
        weight[2] = "geek";
        weight[3] = "gekrt";
        weight[4] = "tree";
        weight[5] = "eetr";
        weight[6] = "egek";
  
        // Edges of the tree
        graph[1].Add(2);
        graph[2].Add(3);
        graph[2].Add(4);
        graph[1].Add(5);
        graph[5].Add(6);
  
        dfs(1, 1);
  
        Console.Write(cnt);
    }
}
  
// This code is contributed by PrinciRaj1992

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Output:

2

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