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.

Implementation:

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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Python3

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# Python3 implementation of the approach
cnt = 0
  
graph = [0] * 100
for i in range(100):
    graph[i] = []
  
weight = ["0"] * 100
  
# Function that returns true
# if x is a palindrome
def isPalindrome(x):
    n = len(x)
  
    for i in range(0, n // 2):
        if x[i] != x[n - 1 - i]:
            return False
  
    return True
  
# Function to perform dfs
def dfs(node, parent):
    global cnt
  
    # Weight of the current node
    x = weight[node]
  
    # If the weight is a palindrome
    if (isPalindrome(x)):
        cnt += 1
  
    for to in graph[node]:
        if to == parent:
            continue
        dfs(to, node)
  
# Driver Code
if __name__ == "__main__":
  
    # Weights of the node
    weight[0] = ""
    weight[1] = "abc"
    weight[2] = "aba"
    weight[3] = "bcb"
    weight[4] = "moh"
    weight[5] = "aa"
  
    # 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
# sanjeev2552

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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

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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