Cuente los Nodes de un árbol cuya string ponderada es un anagrama de la string dada

Dado un árbol y los pesos (en forma de strings) de todos los Nodes, la tarea es contar los Nodes cuya string ponderada es un anagrama con la string dada str .
Ejemplos: 
 

Aporte: 
 

str = “geek” 
Salida: 2 
Solo las strings ponderadas de los Nodes 2 y 6 
son anagramas de la string dada “geek”. 
 

Enfoque: Realice dfs en el árbol y para cada Node, verifique si su string ponderada es un anagrama con la string dada o no. De lo contrario, incremente el conteo.
A continuación se muestra la implementación del enfoque anterior:
 

// 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;
}

// 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

# 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

// 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

<script>
  
// Javascript implementation of the approach
let s;
let cnt = 0;
 
let graph = new Array(100);
let weight = new Array(100);
for(let i = 0; i < 100; i++)
{
    graph[i] = [];
    weight[i] = 0;
}
 
const sort1 = str => str.split('').sort((a, b) => a.localeCompare(b)).join('');
 
// Function that return true if both
// the strings are anagram of each other
function anagram(x, s)
{
    x = sort1(x);
    s = sort1(s);
  
    if (x == s)
        return true;
    else
        return false;
}
 
// Function to perform dfs
function dfs(node, parent)
{
    // If current node's weighted
    // string is an anagram of
    // the given string s
    if (anagram(weight[node], s))
        cnt += 1;
 
    for(let to = 0; to < graph[node].length; to++)
    {
        if(graph[node][to] == parent)
            continue
        dfs(graph[node][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].push(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);
    graph[5].push(6);
 
    dfs(1, 1);
    document.write(cnt);
 
    // This code is contributed by Dharanendra L V.
      
</script>

2

Análisis de Complejidad: 
 

• Complejidad de tiempo: O(N*(S*log(S))). 
  En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Además, para procesar cada Node se usa la función sort() que tiene una complejidad de O(S*log(S)) donde S es la longitud de la string ponderada. Por lo tanto, la complejidad temporal es O(N*(S*log(S))) donde S es la longitud máxima de la string de peso en el árbol.
• Espacio Auxiliar : O(1). 
  No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.