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

Last Updated : 06 Mar, 2023

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>

Output:

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.