Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights do not contain any duplicate character.
Examples:
Input:
Output: 2
Only the strings of the node 1 and 4 contains unique strings.
Approach: Perform dfs on the tree and for every node, check if its string contains duplicate char 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; int cnt = 0; vector< int > graph[100]; vector<string> weight(100); // Function that returns true if the // string contains unique characters bool uniqueChars(string x) { unordered_map< char , int > mp; int n=x.size(); for ( int i = 0; i < n; i++){ mp[x[i]]++; //if more than 1 time present a character if (mp[x[i]]>1) return false ; } return true ; } // Function to perform dfs void dfs( int node) { // If weighted string of the current // node contains unique characters if (uniqueChars(weight[node])) cnt += 1; for ( int to : graph[node]) { dfs(to); } } // Driver code int main() { // Weights of the nodes 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); 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 the // String contains unique characters static boolean uniqueChars( char [] arr) { HashMap<Character, Integer> mp = new HashMap<Character, Integer>(); int n = arr.length; for ( int i = 0 ; i < n; i++){ if (mp.containsKey(arr[i])) { return false ; } else { mp.put(arr[i], 1 ); } } return true ; } // Function to perform dfs static void dfs( int node) { // If weighted String of the current // node contains unique characters if (uniqueChars(weight[node].toCharArray())) cnt += 1 ; for ( int to : graph[node]) { dfs(to); } } // Driver code public static void main(String[] args) { for ( int i = 0 ; i < 100 ; i++) graph[i] = new Vector<Integer>(); // Weights of the nodes weight[ 1 ] = "abc" ; weight[ 2 ] = "aba" ; weight[ 3 ] = "bcb" ; weight[ 4 ] = "moh" ; weight[ 5 ] = "aa" ; // Edges of the tree graph[ 1 ].add( 2 ); graph[ 2 ].add( 3 ); graph[ 2 ].add( 4 ); graph[ 1 ].add( 5 ); dfs( 1 ); System.out.print(cnt); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach cnt = 0 graph = [[] for i in range ( 100 )] weight = [ 0 ] * 100 # Function that returns true if the # string contains unique characters def uniqueChars(x): mp = {} n = len (x) for i in range (n): if x[i] not in mp: mp[x[i]] = 0 else : return False mp[x[i]] + = 1 return True # Function to perform dfs def dfs(node): global cnt, x # If weight of the current node # node contains unique characters if (uniqueChars(weight[node])): cnt + = 1 for to in graph[node]: dfs(to) # Driver code x = 5 # 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 ].append( 2 ) graph[ 2 ].append( 3 ) graph[ 2 ].append( 4 ) graph[ 1 ].append( 5 ) dfs( 1 ) print (cnt) # This code is contributed by SHUBHAMSINGH10 |
C#
// 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 the // String contains unique characters static bool uniqueChars( char [] arr) { Dictionary< char , int > mp = new Dictionary< char , int >(); int n = arr.Length; for ( int i = 0; i < n; i++) if (mp.ContainsKey(arr[i])) { return false ; } else { mp.Add(arr[i], 1); } return true ; } // Function to perform dfs static void dfs( int node) { // If weighted String of the current // node contains unique characters if (uniqueChars(weight[node].ToCharArray())) cnt += 1; foreach ( int to in graph[node]) { dfs(to); } } // Driver code public static void Main(String[] args) { for ( int i = 0; i < 100; i++) graph[i] = new List< int >(); // Weights of the nodes weight[1] = "abc" ; weight[2] = "aba" ; weight[3] = "bcb" ; weight[4] = "moh" ; weight[5] = "aa" ; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1); Console.Write(cnt); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript implementation of the approach let cnt = 0; let graph = new Array(); for (let i = 0; i < 100; i++) { graph.push([]) } let weight = new Array(100); // Function that returns true if the // string contains unique characters function uniqueChars(x) { let mp = new Map(); let n = x.length; for (let i = 0; i < n; i++) { if (mp.has(x[i])) { return false ; } else { mp.set(x[i], 1) } } return true ; } // Function to perform dfs function dfs(node) { // If weighted string of the current // node contains unique characters if (uniqueChars(weight[node])) cnt += 1; for (let to of graph[node]) { dfs(to); } } // Driver code // Weights of the nodes 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); document.write(cnt); // This code is contributed by _saurabh_jaiswal </script> |
2
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) if there are total N nodes in the tree. Also, processing of every node involves traversing the weighted string of that node thus adding a complexity of O(Len) where Len is the length of the weighted string. Therefore, the 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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