计算加权字符串不包含任何重复字符的树的节点

给定一棵树，以及所有节点的权重（以字符串的形式），任务是计算权重不包含任何重复字符的节点。
例子：

Input: 
 

Output: 2 
Only the strings of the node 1 and 4 contains unique strings. 
 

方法：在树上执行 dfs，对于每个节点，检查它的字符串包含重复的字符，如果没有，则增加计数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int cnt = 0;
 
vector graph[100];
vector weight(100);
 
// Function that returns true if the
// string contains unique characters
bool uniqueChars(string x)
{
    map 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;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    static int cnt = 0;
 
    static Vector[] 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 mp =
        new HashMap();
        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();
         
        // 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

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
        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

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static int cnt = 0;
 
    static List[] graph = new List[100];
    static String[] weight = new String[100];
 
    // Function that returns true if the
    // String contains unique characters
    static bool uniqueChars(char[] arr)
    {
        Dictionary mp =
        new Dictionary();
        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();
         
        // 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

Javascript

 
// 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])) {
            mp.set(x[i], mp.get(x[i]) + 1)
        } else {
            mp.set(x[i], 1)
        }
    }
    if (mp.size == x.length)
        return true;
    else
        return false;
}
 
// Function to perform dfs
function dfs(node, parent) {
    // If weighted string of the current
    // node contains unique characters
    if (uniqueChars(weight[node]))
        cnt += 1;
 
    for (let to of graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// 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, 1);
 
document.write(cnt);
 
// This code is contributed by _saurabh_jaiswal
 

输出：

2

复杂度分析：

• 时间复杂度： O(N*Len) 其中 Len 是给定树中节点的加权字符串的最大长度。
  在 dfs 中，树的每个节点都被处理一次，因此如果树中总共有 N 个节点，则由于 dfs 的复杂性是 O(N)。此外，每个节点的处理都涉及遍历该节点的加权字符串，因此增加了 O(Len) 的复杂度，其中 Len 是加权字符串的长度。因此，时间复杂度为 O(N*Len)。
• 辅助空间： O(1)。
  不需要任何额外的空间，因此空间复杂度是恒定的。

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