Count the nodes of the given tree whose weighted string is a palindrome

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

Examples:

Input:

Output: 3
Only the weights of the nodes 2, 3 and 5 are palindromes.

Approach: Perform dfs on the tree and for every node, check if it’s string is palindrome or not. 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 x is a palindrome
bool isPalindrome(string x)
{
    int n = x.size();
    for (int i = 0; i < n / 2; i++) {
        if (x[i] != x[n - 1 - i])
            return false;
    }
    return true;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
  
    // Weight of the current node
    string x = weight[node];
  
    // If the weight is a palindrome
    if (isPalindrome(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] = "abc";
    weight[2] = "aba";
    weight[3] = "bcb";
    weight[4] = "moh";
    weight[5] = "aa";
  
    // 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 x is a palindrome 
static boolean isPalindrome(String x) 
    int n = x.length(); 
    for (int i = 0; i < n / 2; i++)
    
        if (x.charAt(i) != x.charAt(n - 1 - i)) 
            return false
    
    return true
  
// Function to perform dfs 
static void dfs(int node, int parent) 
  
    // Weight of the current node 
    String x = weight.get(node); 
      
  
    // If the weight is a palindrome 
    if (isPalindrome(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( "abc"); 
    weight.add( "aba"); 
    weight.add( "bcb"); 
    weight.add( "moh"); 
    weight.add( "aa"); 
      
    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 Arnab Kundu

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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 x is a palindrome
    static bool isPalindrome(string x)
    {
        int n = x.Length;
        for (int i = 0; i < n / 2; i++)
        {
            if (x[i] != x[n - 1 - i])
                return false;
        }
        return true;
    }
      
    // Function to perform dfs 
    static void dfs(int node, int parent) 
    
      
        // Weight of the current node 
        String x = weight[node]; 
      
        // If the weight is a palindrome 
        if (isPalindrome(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( "abc"); 
        weight.Add( "aba"); 
        weight.Add( "bcb"); 
        weight.Add( "moh"); 
        weight.Add( "aa"); 
          
        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:

3


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Data science |Machine learning|Programming

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