计算与子树节点连接时构成 pangram 的树节点

给定一棵树，以及所有节点的权重（以字符串的形式），任务是计算当与子树节点的字符串连接时加权字符串成为全字母组合的节点。
Pangram： pangram 是包含英文字母表中每个字母的句子。

例子：

Input: 
 

Output: 1 
Only the weighted string of sub-tree of node 1 makes the pangram. 

方法：对树执行 dfs 并更新每个节点的权重，使其权重与子树节点的权重串联存储。然后，计算更新后的加权字符串形成全词组的节点。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
vector graph[100];
vector weight(100);
 
// Function that returns if the
// string x is a pangram
bool Pangram(string x)
{
    map mp;
    int n = x.size();
 
    for (int i = 0; i < n; i++)
        mp[x[i]]++;
    if (mp.size() == 26)
        return true;
    else
        return false;
}
 
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
int countTotalPangram(int n)
{
    int cnt = 0;
    for (int i = 1; i <= n; i++)
        if (Pangram(weight[i]))
            cnt++;
    return cnt;
}
 
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
void dfs(int node, int parent)
{
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
        weight[node] += weight[to];
    }
}
 
// Driver code
int main()
{
    int n = 6;
 
    // Weights of the nodes
    weight[1] = "abcde";
    weight[2] = "fghijkl";
    weight[3] = "abcdefg";
    weight[4] = "mnopqr";
    weight[5] = "stuvwxy";
    weight[6] = "zabcdef";
 
    // 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 << countTotalPangram(n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG{
     
@SuppressWarnings("unchecked")
static Vector []graph = new Vector[100];
static String []weight = new String[100];
 
// Function that returns if the
// String x is a pangram
static boolean Pangram(String x)
{
    HashMap mp = new HashMap<>();
    int n = x.length();
 
    for(int i = 0 ; i < n; i++)
    {
        if (mp.containsKey(x.charAt(i)))
        {
            mp.put(x.charAt(i),
            mp.get(x.charAt(i)) + 1);
        }
        else
        {
            mp.put(x.charAt(i), 1);
        }
    }
    if (mp.size() == 26)
        return true;
    else
        return false;
}
 
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
static int countTotalPangram(int n)
{
    int cnt = 0;
    for(int i = 1; i <= n; i++)
        if (Pangram(weight[i]))
            cnt++;
             
    return cnt;
}
 
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
static void dfs(int node, int parent)
{
    for(int to : graph[node])
    {
        if (to == parent)
            continue;
             
        dfs(to, node);
        weight[node] += weight[to];
    }
}
 
// Driver code
public static void main(String[] args)
{
    int n = 6;
 
    // Weights of the nodes
    weight[1] = "abcde";
    weight[2] = "fghijkl";
    weight[3] = "abcdefg";
    weight[4] = "mnopqr";
    weight[5] = "stuvwxy";
    weight[6] = "zabcdef";
 
    for(int i = 0; i < graph.length; i++)
        graph[i] = new Vector();
         
    // 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(countTotalPangram(n));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 implementation of the approach
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function that returns if the
# string x is a pangram
def Pangram(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)== 26):
        return True
    else:
        return False
 
# Function to return the count of nodes
# which make pangram with the
# sub-tree nodes
def countTotalPangram(n):
    cnt = 0
    for i in range(1, n + 1):
        if (Pangram(weight[i])):
            cnt += 1
    return cnt
 
# Function to perform dfs and update the nodes
# such that weight[i] will store the weight[i]
# concatenated with the weights of
# all the nodes in the sub-tree
def dfs(node, parent):
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
        weight[node] += weight[to]
 
# Driver code
n = 6
 
# Weights of the nodes
weight[1] = "abcde"
weight[2] = "fghijkl"
weight[3] = "abcdefg"
weight[4] = "mnopqr"
weight[5] = "stuvwxy"
weight[6] = "zabcdef"
 
# 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(countTotalPangram(n))
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of
// the above approach
using System;
using System.Collections.Generic;
class GFG{   
 
static List []graph =
            new List[100];
static String []weight =
                new String[100];
 
// Function that returns if the
// String x is a pangram
static bool Pangram(String x)
{
  Dictionary mp = new Dictionary();
  int n = x.Length;
 
  for(int i = 0 ; i < n; i++)
  {
    if (mp.ContainsKey(x[i]))
    {
      mp[x[i]] = mp[x[i]] + 1;
    }
    else
    {
      mp.Add(x[i], 1);
    }
  }
  if (mp.Count == 26)
    return true;
  else
    return false;
}
 
// Function to return the
// count of nodes which
// make pangram with the
// sub-tree nodes
static int countTotalPangram(int n)
{
  int cnt = 0;
  for(int i = 1; i <= n; i++)
    if (Pangram(weight[i]))
      cnt++;
 
  return cnt;
}
 
// Function to perform dfs and
// update the nodes such that
// weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
static void dfs(int node, int parent)
{
  foreach(int to in graph[node])
  {
    if (to == parent)
      continue;
 
    dfs(to, node);
    weight[node] += weight[to];
  }
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 6;
 
  // Weights of the nodes
  weight[1] = "abcde";
  weight[2] = "fghijkl";
  weight[3] = "abcdefg";
  weight[4] = "mnopqr";
  weight[5] = "stuvwxy";
  weight[6] = "zabcdef";
 
  for(int i = 0;
          i < graph.Length; i++)
    graph[i] = new List();
 
  // 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(countTotalPangram(n));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// JavaScript implementation of the approach
 
 
let graph = new Array();
 
for (let i = 0; i < 100; i++) {
    graph.push([])
}
let weight = new Array(100).fill(0);
 
// Function that returns if the
// string x is a pangram
function Pangram(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 == 26)
        return true;
    else
        return false;
}
 
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
function countTotalPangram(n) {
    let cnt = 0;
    for (let i = 1; i <= n; i++)
        if (Pangram(weight[i]))
            cnt++;
    return cnt;
}
 
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
function dfs(node, parent) {
 
    for (let to of graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
        weight[node] += weight[to];
    }
}
 
// Driver code
 
let n = 6;
 
// Weights of the nodes
weight[1] = "abcde";
weight[2] = "fghijkl";
weight[3] = "abcdefg";
weight[4] = "mnopqr";
weight[5] = "stuvwxy";
weight[6] = "zabcdef";
 
// 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(countTotalPangram(n));
 
 
// This code is contributed by gfgking
 

输出：

1

复杂度分析：

• 时间复杂度： O(N*S)。
  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，为了处理每个节点，Pangram()函数用于每个复杂度为 O(S) 的节点，其中 S 是子树中所有权重字符串的长度之和，因为这是对每个节点完成的，所以这部分的整体时间复杂度变为 O(N*S)。因此，最终的时间复杂度为 O(N*S)。
• 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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