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

Given an N-ary tree, and the weights which are in the form of strings of all the nodes, the task is to count the number of leaf nodes whose weights are palindrome.

Examples:

Input: 
               1(ab)
              /  \
       (abca)2     5 (aba)
          /   \ 
   (axxa)3     4 (geeks)
Output: 2
Explanation: 
Only the weights of the leaf nodes
"axxa" and "aba" are palindromes.

Input: 
               1(abx)
              /  
            2(abaa) 
           /    
          3(amma)  
Output: 1
Explanation: 
Only the weight of the leaf
node "amma" is palindrome.

Approach: To solve the problem mentioned above follow the steps given below:

Depth First Search can be used to traverse the complete tree.
We will keep track of parent while traversing to avoid the visited node array.
Initially for every node we can set a flag and if the node have at least one child (i.e. non-leaf node) then we will reset the flag.
The nodes with no children are the leaf nodes. For every leaf node, we will check if it’s string is palindrome or not. If yes then increment the count.

Below is the implementation of the above approach:

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 on the tree

void dfs(int node, int parent)
{

    int flag = 1;
 
    // Iterating the children of current node

    for (int to : graph[node]) {
 
        // There is at least a child

        // of the current node

        if (to == parent)

            continue;

        flag = 0;

        dfs(to, node);

    }
 
    // Current node is connected to only

    // its parent i.e. it is a leaf node

    if (flag == 1) {

        // Weight of the current node

        string x = weight[node];
 
        // If the weight is a palindrome

        if (isPalindrome(x))

            cnt += 1;

    }
}
 
// Driver code

int main()
{
 
    // Weights of the node

    weight[1] = "ab";

    weight[2] = "abca";

    weight[3] = "axxa";

    weight[4] = "geeks";

    weight[5] = "aba";
 
    // 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<Integer> []graph = new Vector[100];

static String []weight = new String[100];
 
// 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 on the tree

static void dfs(int node, int parent)
{

    int flag = 1;
 
    // Iterating the children of current node

    for (int to : graph[node]) 

    {
 
        // There is at least a child

        // of the current node

        if (to == parent)

            continue;

        flag = 0;

        dfs(to, node);

    }
 
    // Current node is connected to only

    // its parent i.e. it is a leaf node

    if (flag == 1) 

    {

        // Weight of the current node

        String x = weight[node];
 
        // If the weight is a palindrome

        if (isPalindrome(x))

            cnt += 1;

    }
}
 
// Driver code

public static void main(String[] args)
{

    for(int i = 0; i < graph.length;i++)

        graph[i] = new Vector<Integer>();

    // Weights of the node

    weight[1] = "ab";

    weight[2] = "abca";

    weight[3] = "axxa";

    weight[4] = "geeks";

    weight[5] = "aba";
 
    // Edges of the tree

    graph[1].add(2);

    graph[2].add(3);

    graph[2].add(4);

    graph[1].add(5);
 
    dfs(1, 1);
 
    System.out.print(cnt);
}
}
 
// This code is contributed by amal kumar choubey

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: str) -> bool:

    n = len(x)

    for i in range(n // 2):

        if (x[i] != x[n - 1 - i]):

            return False
 
    return True
 
# Function to perform DFS on the tree

def dfs(node: int, parent: int) -> None:

    global cnt, graph, weight
 
    flag = 1
 
    # Iterating the children of current node

    for to in graph[node]:
 
        # There is at least a child

        # of the current node

        if (to == parent):

            continue

        flag = 0

        dfs(to, node)
 
    # Current node is connected to only

    # its parent i.e. it is a leaf node

    if (flag == 1):

        # Weight of the current node

        x = weight[node]
 
        # If the weight is a palindrome

        if (isPalindrome(x)):

            cnt += 1
 
# Driver code

if __name__ == "__main__":
 
    # Weights of the node

    weight[1] = "ab"

    weight[2] = "abca"

    weight[3] = "axxa"

    weight[4] = "geeks"

    weight[5] = "aba"
 
    # 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# implementation of the approach

using System;

using System.Collections.Generic;
 
class GFG{
 
static int cnt = 0;
 
static List<int> []graph = new List<int>[100];

static String []weight = new String[100];
 
// 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 on the tree

static void dfs(int node, int parent)
{

    int flag = 1;
 
    // Iterating the children of 

    // current node

    foreach (int to in graph[node]) 

    {
 
        // There is at least a child

        // of the current node

        if (to == parent)

            continue;

        flag = 0;

        dfs(to, node);

    }
 
    // Current node is connected to only

    // its parent i.e. it is a leaf node

    if (flag == 1) 

    {

        // Weight of the current node

        String x = weight[node];
 
        // If the weight is a palindrome

        if (isPalindrome(x))

            cnt += 1;

    }
}
 
// Driver code

public static void Main(String[] args)
{

    for(int i = 0; i < graph.Length; i++)

       graph[i] = new List<int>();

    // Weights of the node

    weight[1] = "ab";

    weight[2] = "abca";

    weight[3] = "axxa";

    weight[4] = "geeks";

    weight[5] = "aba";
 
    // Edges of the tree

    graph[1].Add(2);

    graph[2].Add(3);

    graph[2].Add(4);

    graph[1].Add(5);
 
    dfs(1, 1);
 
    Console.Write(cnt);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

<script>
 
    // JavaScript implementation of the approach

    let cnt = 0;

    let graph = new Array(100);

    let weight = new Array(100);
 
    // Function that returns true

    // if x is a palindrome

    function isPalindrome(x)

    {

        let n = x.length;

        for (let i = 0; i < parseInt(n / 2, 10); i++)

        {

            if (x[i] != x[n - 1 - i])

                return false;

        }

        return true;

    }
 
    // Function to perform DFS on the tree

    function dfs(node, parent)

    {

        let flag = 1;
 
        // Iterating the children of current node

        for (let to = 0; to < graph[node].length; to++)

        {
 
            // There is at least a child

            // of the current node

            if (graph[node][to] == parent)

                continue;

            flag = 0;

            dfs(graph[node][to], node);

        }
 
        // Current node is connected to only

        // its parent i.e. it is a leaf node

        if (flag == 1)

        {

            // Weight of the current node

            let x = weight[node];
 
            // If the weight is a palindrome

            if (isPalindrome(x))

                cnt += 1;

        }

    }

    for(let i = 0; i < graph.length;i++)

        graph[i] = [];

    // Weights of the node

    weight[1] = "ab";

    weight[2] = "abca";

    weight[3] = "axxa";

    weight[4] = "geeks";

    weight[5] = "aba";

    // 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);

</script>