Count the nodes of a tree whose weighted string does not contain any duplicate characters

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weights does 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 it’s string contains duplicate char or not, If not then increment the count.

Below is the implementation of the above approach:

C++

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// 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)
{
    map<char, int> mp;
    int n = x.size();
  
    for (int i = 0; i < n; i++)
        mp[x[i]]++;
    if (mp.size() == x.size())
        return true;
    else
        return false;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weighted string of the current
    // node contains unique characters
    if (uniqueChars(weight[node]))
        cnt += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// 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, 1);
  
    cout << cnt;
  
    return 0;
}

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Java

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// 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])) 
            {
                mp.put(arr[i], mp.get(arr[i]) + 1);
            }
            else
            {
                mp.put(arr[i], 1);
            }
        if (mp.size() == arr.length)
            return true;
        else
            return false;
    }
  
    // Function to perform dfs
    static void dfs(int node, int parent) 
    {
        // If weighted String of the current
        // node contains unique characters
        if (uniqueChars(weight[node].toCharArray()))
            cnt += 1;
  
        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
  
    // 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, 1);
  
        System.out.print(cnt);
    }
}
  
// This code is contributed by Rajput-Ji

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Python3

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# 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
        mp[x[i]] += 1
    if (len(mp) == len(x)):
        return True
    else:
        return False
  
# Function to perform dfs 
def dfs(node, parent):
    global cnt, x
      
    # If weight of the current node 
    # node contains unique characters 
    if (uniqueChars(weight[node])):
        cnt += 1
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# 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, 1)
print(cnt)
  
# This code is contributed by SHUBHAMSINGH10

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C#

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// 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])) 
            {
                mp[arr[i]] = mp[arr[i]] + 1;
            }
            else
            {
                mp.Add(arr[i], 1);
            }
        if (mp.Count == arr.Length)
            return true;
        else
            return false;
    }
  
    // Function to perform dfs
    static void dfs(int node, int parent) 
    {
        // If weighted String of the current
        // node contains unique characters
        if (uniqueChars(weight[node].ToCharArray()))
            cnt += 1;
  
        foreach (int to in graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
  
    // 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, 1);
  
        Console.Write(cnt);
    }
}
  
// This code is contributed by PrinciRaj1992

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Output:

2



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