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

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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++




// 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


Javascript




<script>
  
// Javascript implementation of the approach
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;
}
 
// Function that returns true
// if x is a palindrome
function isPalindrome(x)
{
    let n = x.length;
    for(let i = 0; i < n / 2; i++)
    {
        if (x[i] != x[n - 1 - i])
            return false;
    }
    return true;
}
 
// Function to perform dfs
function dfs(node, parent)
{
 
    // Weight of the current node
    let x = weight[node];
 
    // If the weight is a palindrome
    if (isPalindrome(x))
        cnt += 1;
         
    for(let to = 0; to < graph[node].length; to++)
    {
        if (graph[node][to] == parent)
            continue
             
        dfs(graph[node][to], node); 
    }
}
 
// Driver code
     
// 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(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
 
dfs(1, 1);
 
document.write(cnt);
 
// This code is contributed by Dharanendra L V.
      
</script>


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.

 New Approach :  (BFS)

  • This implementation uses a queue to perform a breadth-first search starting from the root node (node 1).
  •  It iteratively visits each node and checks if its weight is a palindrome.
  • If it is, the count is incremented.
  • The neighbors of each node are added to the queue for further exploration.
  • Finally, the count of nodes with palindrome weights is returned.

Bellow is the implementation of the approach ; 

C++




#include <iostream>
#include <deque>
#include <unordered_map>
#include <vector>
#include <string>
using namespace std;
 
bool is_palindrome(const string& x) {
    return x == string(x.rbegin(), x.rend());
}
 
int count_palindrome_nodes(const unordered_map<int, vector<int>>& graph, const vector<string>& weight) {
    int count = 0;
    deque<int> queue;
    queue.push_back(1); // Start the BFS from node 1 (root)
 
    while (!queue.empty()) {
        int node = queue.front();
        queue.pop_front();
 
        if (is_palindrome(weight[node])) {
            count++;
        }
 
        for (int neighbor : graph.at(node)) {
            queue.push_back(neighbor);
        }
    }
 
    return count;
}
 
int main() {
    // Weights of the nodes
    vector<string> weight = {"", "abc", "aba", "bcb", "moh", "aa"};
 
    // Edges of the tree
    unordered_map<int, vector<int>> graph = {
        {1, {2, 5}},
        {2, {3, 4}},
        {3, {}},
        {4, {}},
        {5, {}}
    };
 
    int result = count_palindrome_nodes(graph, weight);
    cout << result << endl;
 
    return 0;
}


Java




import java.util.*;
 
public class GFG {
 
    // Function to check if a string is a palindrome
    static boolean isPalindrome(String x)
    {
        return x.equals(
            new StringBuilder(x).reverse().toString());
    }
 
    // Function to count palindrome nodes in the tree
    static int countPalindromeNodes(
        HashMap<Integer, List<Integer> > graph,
        List<String> weight)
    {
        int count = 0;
        Deque<Integer> queue = new LinkedList<>();
        queue.add(1); // Start the BFS from node 1 (root)
 
        while (!queue.isEmpty()) {
            int node = queue.poll();
 
            if (isPalindrome(weight.get(node))) {
                count++;
            }
 
            for (int neighbor : graph.getOrDefault(
                     node, Collections.emptyList())) {
                queue.add(neighbor);
            }
        }
 
        return count;
    }
 
    public static void main(String[] args)
    {
        // Weights of the nodes
        List<String> weight = new ArrayList<>(Arrays.asList(
            "", "abc", "aba", "bcb", "moh", "aa"));
 
        // Edges of the tree
        HashMap<Integer, List<Integer> > graph
            = new HashMap<>();
        graph.put(1, new ArrayList<>(Arrays.asList(2, 5)));
        graph.put(2, new ArrayList<>(Arrays.asList(3, 4)));
        graph.put(3, new ArrayList<>());
        graph.put(4, new ArrayList<>());
        graph.put(5, new ArrayList<>());
 
        int result = countPalindromeNodes(graph, weight);
        System.out.println(result);
    }
}


Python




from collections import deque
 
def is_palindrome(x):
    return x == x[::-1]
 
def count_palindrome_nodes(graph, weight):
    count = 0
    queue = deque()
    queue.append(1# Start the BFS from node 1 (root)
 
    while queue:
        node = queue.popleft()
        if is_palindrome(weight[node]):
            count += 1
 
        for neighbor in graph[node]:
            queue.append(neighbor)
 
    return count
 
# Example usage:
 
# Weights of the nodes
weight = ["", "abc", "aba", "bcb", "moh", "aa"]
 
# Edges of the tree
graph = {
    1: [2, 5],
    2: [3, 4],
    3: [],
    4: [],
    5: []
}
 
result = count_palindrome_nodes(graph, weight)
print(result)


C#




using System;
using System.Collections.Generic;
 
class GFG
{
    static bool IsPalindrome(string x)
    {
        char[] charArray = x.ToCharArray();
        Array.Reverse(charArray);
        string reversed = new string(charArray);
        return x == reversed;
    }
 
    static int CountPalindromeNodes(Dictionary<int, List<int>> graph, List<string> weight)
    {
        int count = 0;
        Queue<int> queue = new Queue<int>();
        queue.Enqueue(1); // Start the BFS from node 1 (root)
 
        while (queue.Count > 0)
        {
            int node = queue.Dequeue();
 
            if (IsPalindrome(weight[node]))
            {
                count++;
            }
 
            foreach (int neighbor in graph[node])
            {
                queue.Enqueue(neighbor);
            }
        }
 
        return count;
    }
 
    static void Main(string[] args)
    {
        // Weights of the nodes
        List<string> weight = new List<string> { "", "abc", "aba", "bcb", "moh", "aa" };
 
        // Edges of the tree
        Dictionary<int, List<int>> graph = new Dictionary<int, List<int>>
        {
            { 1, new List<int> { 2, 5 } },
            { 2, new List<int> { 3, 4 } },
            { 3, new List<int>() },
            { 4, new List<int>() },
            { 5, new List<int>() }
        };
 
        int result = CountPalindromeNodes(graph, weight);
        Console.WriteLine(result);
    }
}


Javascript




// Function to check if a string is a palindrome
function is_palindrome(x) {
    return x === x.split("").reverse().join("");
}
 
 // Function to count palindrome nodes in the tree
function count_palindrome_nodes(graph, weight) {
    let count = 0;
    let queue = [];
    queue.push(1);  // Start the BFS from node 1 (root)
 
    while (queue.length > 0) {
        let node = queue.shift();
 
        if (is_palindrome(weight[node])) {
            count++;
        }
 
        let neighbors = graph[node];
        for (let neighbor of neighbors) {
            queue.push(neighbor);
        }
    }
 
    return count;
}
 
let weight = {
    1: "abc",
    2: "aba",
    3: "bcb",
    4: "moh",
    5: "aa"
};
 
// Edges of the tree
let graph = {
    1: [2, 5],
    2: [3, 4],
    3: [],
    4: [],
    5: []
};
 
let result = count_palindrome_nodes(graph, weight);
console.log(result);
 
// This Code Is Contributed By Shubham Tiwari


Output

3

Time Complexity: O(N)

Auxiliary Space: O(N)



Last Updated : 13 Sep, 2023
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