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++
// 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 dfsvoid 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 codeint 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 approachcnt = 0 graph = [0] * 100for i in range(100): graph[i] = [] weight = ["0"] * 100 # Function that returns true# if x is a palindromedef 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 dfsdef 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 Codeif __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:
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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