Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weighted string is an anagram with the given string str.
Examples:
Input:
str = “geek”
Output: 2
Only the weighted strings of the nodes 2 and 6
are anagrams of the given string “geek”.
Approach: Perform dfs on the tree and for every node, check if it’s weighted string is anagram with the given string or not, If not 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;string s;int cnt = 0;vector<int> graph[100];vector<string> weight(100);// Function that return true if both// the strings are anagram of each otherbool anagram(string x, string s){ sort(x.begin(), x.end()); sort(s.begin(), s.end()); if (x == s) return true; else return false;}// Function to perform dfsvoid dfs(int node, int parent){ // If current node's weighted // string is an anagram of // the given string s if (anagram(weight[node], s)) cnt += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); }}// Driver codeint main(){ s = "geek"; // Weights of the nodes weight[1] = "eeggk"; weight[2] = "geek"; weight[3] = "gekrt"; weight[4] = "tree"; weight[5] = "eetr"; weight[6] = "egek"; // Edges of the tree graph[1].push_back(2); graph[2].push_back(3); graph[2].push_back(4); graph[1].push_back(5); graph[5].push_back(6); dfs(1, 1); cout << cnt; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { static String s; static int cnt = 0; static Vector<Integer>[] graph = new Vector[100]; static String[] weight = new String[100]; // Function that return true if both // the Strings are anagram of each other static boolean anagram(String x, String s) { x = sort(x); s = sort(s); if (x.equals(s)) return true; else return false; } static String sort(String inputString) { // convert input string to char array char tempArray[] = inputString.toCharArray(); // sort tempArray Arrays.sort(tempArray); // return new sorted string return new String(tempArray); } // Function to perform dfs static void dfs(int node, int parent) { // If current node's weighted // String is an anagram of // the given String s if (anagram(weight[node], s)) cnt += 1; for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code public static void main(String[] args) { s = "geek"; for (int i = 0; i < 100; i++) graph[i] = new Vector<Integer>(); // Weights of the nodes weight[1] = "eeggk"; weight[2] = "geek"; weight[3] = "gekrt"; weight[4] = "tree"; weight[5] = "eetr"; weight[6] = "egek"; // Edges of the tree graph[1].add(2); graph[2].add(3); graph[2].add(4); graph[1].add(5); graph[5].add(6); dfs(1, 1); System.out.print(cnt); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach cnt = 0graph = [[] for i in range(100)]weight = [0] * 100# Function that return true if both # the strings are anagram of each other def anagram(x, s): x = sorted(list(x)) s = sorted(list(s)) if (x == s): return True else: return False# Function to perform dfs def dfs(node, parent): global cnt, s # If weight of the current node # string is an anagram of # the given string s if (anagram(weight[node], s)): cnt += 1 for to in graph[node]: if (to == parent): continue dfs(to, node)# Driver code s = "geek"# Weights of the nodes weight[1] = "eeggk"weight[2] = "geek"weight[3] = "gekrt"weight[4] = "tree"weight[5] = "eetr"weight[6] = "egek"# Edges of the tree graph[1].append(2)graph[2].append(3)graph[2].append(4)graph[1].append(5)graph[5].append(6)dfs(1, 1)print(cnt)# This code is contributed by SHUBHAMSINGH10 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG { static String s; static int cnt = 0; static List<int>[] graph = new List<int>[100]; static String[] weight = new String[100]; // Function that return true if both // the Strings are anagram of each other static bool anagram(String x, String s) { x = sort(x); s = sort(s); if (x.Equals(s)) return true; else return false; } static String sort(String inputString) { // convert input string to char array char []tempArray = inputString.ToCharArray(); // sort tempArray Array.Sort(tempArray); // return new sorted string return new String(tempArray); } // Function to perform dfs static void dfs(int node, int parent) { // If current node's weighted // String is an anagram of // the given String s if (anagram(weight[node], s)) cnt += 1; foreach (int to in graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code public static void Main(String[] args) { s = "geek"; for (int i = 0; i < 100; i++) graph[i] = new List<int>(); // Weights of the nodes weight[1] = "eeggk"; weight[2] = "geek"; weight[3] = "gekrt"; weight[4] = "tree"; weight[5] = "eetr"; weight[6] = "egek"; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); graph[5].Add(6); dfs(1, 1); Console.Write(cnt); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript implementation of the approachlet s;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;}const sort1 = str => str.split('').sort((a, b) => a.localeCompare(b)).join('');// Function that return true if both// the strings are anagram of each other function anagram(x, s){ x = sort1(x); s = sort1(s); if (x == s) return true; else return false;}// Function to perform dfsfunction dfs(node, parent){ // If current node's weighted // string is an anagram of // the given string s if (anagram(weight[node], s)) cnt += 1; for(let to = 0; to < graph[node].length; to++) { if(graph[node][to] == parent) continue dfs(graph[node][to], node); }}// Driver code s = "geek"; // Weights of the nodes weight[1] = "eeggk"; weight[2] = "geek"; weight[3] = "gekrt"; weight[4] = "tree"; weight[5] = "eetr"; weight[6] = "egek"; // Edges of the tree graph[1].push(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); graph[5].push(6); dfs(1, 1); document.write(cnt); // This code is contributed by Dharanendra L V. </script> |
Output:
2
Complexity Analysis:
- Time Complexity : O(N*(S*log(S))).
In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, for processing each node the sort() function is used which has a complexity of O(S*log(S)) where S is the length of the weighted string. Therefore, the time complexity is O(N*(S*log(S))) where S is the maximum length of the weight string in the tree. - Auxiliary Space : O(1).
Any extra space is not required, so the space complexity is constant.
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