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 (neveropen)
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 palindromebool 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 treevoid 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 codeint main(){ // Weights of the node weight[1] = "ab"; weight[2] = "abca"; weight[3] = "axxa"; weight[4] = "neveropen"; 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 approachimport 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 palindromestatic 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 treestatic 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 codepublic 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] = "neveropen"; 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 approachcnt = 0graph = [0] * 100for i in range(100): graph[i] = [] weight = [0] * 100# Function that returns true# if x is a palindromedef 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 treedef 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 codeif __name__ == "__main__": # Weights of the node weight[1] = "ab" weight[2] = "abca" weight[3] = "axxa" weight[4] = "neveropen" 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#
// C# implementation of the approachusing 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 palindromestatic 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 treestatic 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 codepublic 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] = "neveropen"; 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] = "neveropen"; 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> |
Output:
2
Time Complexity: O(N) where N is the number of nodes in the tree.
Auxiliary Space: O(1) as constant space is considered.
