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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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 

Python3

# 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 

C#

// 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 

Output:

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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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