找出给定句子中字谜最多的单词

给定一个句子形式的字符串S ，任务是从文本中找到具有最大字谜数量的单词 出现在给定的句子中。

例子：

Input: S = “please be silent and listen to what the professor says” 
Output: silent 
Explanation: 
Only the word “silent” has an anagram(listen) present in the sentence. 

Input: S = “cat is not a dog and sword has no words when government creates act so what is tac” 
Output: cat 
Explanation: 
The word “cat” has two anagrams (“act”, “tac”) present in the sentence. 
The word “words” has an anagram (“sword”) present in the sentence. 
Hence, the word with the maximum anagrams is “cat”. 

方法：
解决该程序需要进行以下观察：

Property of Prime Numbers: The product of any combination of prime numbers generates an unique number which cannot be obtained by any other combination of prime numbers. 
 

请按照以下步骤解决问题：

• 为每个字母表分配一个不同的质数。
• 对于给定字符串中的每个单词，计算分配给该单词字符的素数的乘积并将其存储在HashMap 中。
• 计算分配给单词字符的素数的乘积并将其存储在HashMap 中。
• 在HashMap 中找到频率最大的产品，并打印相应的字谜之一作为答案。

下面是上述方法的实现：

C++14

// C++ Program to find the word
// with most anagrams in a sentence
#include 
using namespace std;
 
// Function to find the word with
// maximum number of anagrams
string largestAnagramGrp(vector arr)
{
   
  // Primes assigned to 26 alphabets
  int prime[] = {2,  3,  5,  7,  11, 13, 17,
                19, 23, 29, 31, 37, 41,
                43, 47, 53, 59, 61, 67, 71,
                73, 79, 83, 89, 97, 101};
  int max = -1;
  long maxpdt = -1;
 
  // Stores the product and
  // word mappings
  unordered_map W;
 
  // Stores the frequencies
  // of products
  unordered_map P;
  for (string temp : arr)
  {
    long pdt = 1;
 
    // Calculate the product of
    // primes assigned
    for (char t : temp)
    {
      pdt *= prime[t - 'a'];
    }
 
    // If product already exists
    if (P.find(pdt) != P.end())
    {
      P[pdt]++;
    }
 
    // Otherwise
    else
    {
      W[pdt] = temp;
      P[pdt] = 1;
    }
  }
 
  // Fetch the most frequent product
  for (auto e : P)
  {
    if (max < e.second)
    {
      max = e.second;
      maxpdt = e.first;
    }
  }
 
  // Return a string
  // with that product
  return W[maxpdt];
}
 
// Driver Code
 
int main()
{
  char S[] = "please be silent and listen to what the professor says ";
  vector arr;
 
  char *token = strtok(S, " ");
  while (token != NULL)
  {
    arr.push_back(token);
    token = strtok(NULL, " ");
  }
 
  cout << largestAnagramGrp(arr) << endl;
  return 0;
}
 
// This code is contributed by
// sanjeev2552

Java

// Java Program to find the word
// with most anagrams in a sentence
import java.util.*;
 
class GFG {
 
    // Function to find the word with
    // maximum number of anagrams
    private static String largestAnagramGrp(
        String arr[])
    {
        // Primes assigned to 26 alphabets
        int prime[]
            = { 2, 3, 5, 7, 11, 13, 17,
                19, 23, 29, 31, 37, 41,
                43, 47, 53, 59, 61, 67, 71,
                73, 79, 83, 89, 97, 101 };
 
        int max = -1;
        long maxpdt = -1;
 
        // Stores the product and
        // word mappings
        HashMap W
            = new HashMap<>();
 
        // Stores the frequencies
        // of products
        HashMap P
            = new HashMap<>();
 
        for (String temp : arr) {
            char c[] = temp.toCharArray();
            long pdt = 1;
 
            // Calculate the product of
            // primes assigned
            for (char t : c) {
                pdt *= prime[t - 'a'];
            }
 
            // If product already exists
            if (P.containsKey(pdt)) {
                P.put(pdt, P.get(pdt) + 1);
            }
 
            // Otherwise
            else {
                W.put(pdt, temp);
                P.put(pdt, 1);
            }
        }
 
        // Fetch the most frequent product
        for (Map.Entry
                 e : P.entrySet()) {
            if (max < e.getValue()) {
                max = e.getValue();
                maxpdt = e.getKey();
            }
        }
 
        // Return a string
        // with that product
        return W.get(maxpdt);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String S = "please be silent and listen"
                   + " to what the professor says ";
        String arr[] = S.split("[ ]+");
        System.out.println(largestAnagramGrp(arr));
    }
}

Python3

# Python3 Program to find the word
# with most anagrams in a sentence
 
# Function to find the word with
# maximum number of anagrams
def largestAnagramGrp(arr):
     
    # Primes assigned to 26 alphabets
    prime= [2, 3, 5, 7, 11, 13, 17,
            19, 23, 29, 31, 37, 41,
            43, 47, 53, 59, 61, 67, 71,
            73, 79, 83, 89, 97, 101]
    max = -1
    maxpdt = -1
 
    # Stores the product and
    # word mappings
    W = {}
 
    # Stores the frequencies
    # of products
    P = {}
    for temp in arr:
        c = [i for i in temp]
        pdt = 1
 
        # Calculate the product of
        # primes assigned
        for t in c:
            pdt *= prime[ord(t) - ord('a')]
 
        # If product already exists
        if (pdt in P):
            P[pdt] = P.get(pdt, 0) + 1
 
        # Otherwise
        else:
            W[pdt] = temp
            P[pdt] = 1
 
    # Fetch the most frequent product
    for e in P:
        if (max < P[e]):
            max = P[e]
            maxpdt = e
 
    # Return a string
    # with that product
    return W[maxpdt]
 
# Driver Code
if __name__ == '__main__':
    S = "please be silent and listen to what the professor says"
    arr = S.split(" ")
    print(largestAnagramGrp(arr))
 
    # This code is contributed by mohit kumar 29

C#

// C# program to find the word
// with most anagrams in a sentence
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the word with
// maximum number of anagrams
private static String largestAnagramGrp(String []arr)
{
     
    // Primes assigned to 26 alphabets
    int []prime = { 2, 3, 5, 7, 11, 13, 17,
                    19, 23, 29, 31, 37, 41,
                    43, 47, 53, 59, 61, 67, 71,
                    73, 79, 83, 89, 97, 101 };
 
    int max = -1;
    long maxpdt = -1;
 
    // Stores the product and
    // word mappings
    Dictionary W = new Dictionary();
 
    // Stores the frequencies
    // of products
    Dictionary P = new Dictionary();
                                        
    foreach(String temp in arr)
    {
        char []c = temp.ToCharArray();
        long pdt = 1;
 
        // Calculate the product of
        // primes assigned
        foreach(char t in c)
        {
            pdt *= prime[t - 'a'];
        }
 
        // If product already exists
        if (P.ContainsKey(pdt))
        {
            P[pdt] = P[pdt] + 1;
        }
 
        // Otherwise
        else
        {
            W.Add(pdt, temp);
            P.Add(pdt, 1);
        }
    }
 
    // Fetch the most frequent product
    foreach(KeyValuePair e in P)
    {
        if (max < e.Value)
        {
            max = e.Value;
            maxpdt = e.Key;
        }
    }
 
    // Return a string
    // with that product
    return W[maxpdt];
}
 
// Driver Code
public static void Main(String []args)
{
    String S = "please be silent and listen" +
               " to what the professor says ";
                
    String []arr = S.Split(' ');
     
    Console.WriteLine(largestAnagramGrp(arr));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// JavaScript program to find the word
// with most anagrams in a sentence
// Function to find the word with
// maximum number of anagrams
function largestAnagramGrp(arr)
{
     
    // Primes assigned to 26 alphabets
    var prime = [ 2, 3, 5, 7, 11, 13, 17, 19,
                  23, 29, 31, 37, 41, 43, 47,
                  53, 59, 61, 67, 71, 73, 79,
                  83, 89, 97, 101, ];
     
    var max = -1;
    var maxpdt = -1;
     
    // Stores the product and
    // word mappings
    var W = {};
     
    // Stores the frequencies
    // of products
    var P = {};
     
    for (const temp of arr)
    {
        var c = temp.split("");
        var pdt = 1;
         
        // Calculate the product of
        // primes assigned
        for(const t of c)
        {
            pdt *= prime[t.charCodeAt(0) -
                       "a".charCodeAt(0)];
        }
         
        // If product already exists
        if (P.hasOwnProperty(pdt))
        {
            P[pdt] = P[pdt] + 1;
        }
         
        // Otherwise
        else
        {
            W[pdt] = temp;
            P[pdt] = 1;
        }
    }
     
    // Fetch the most frequent product
    for (const [key, value] of Object.entries(P))
    {
        if (max < value)
        {
            max = value;
            maxpdt = key;
        }
    }
     
    // Return a string
    // with that product
    return W[maxpdt];
}
 
// Driver Code
var S = "please be silent and listen" +
        " to what the professor says ";
var arr = S.split(" ");
 
document.write(largestAnagramGrp(arr));
 
// This code is contributed by rdtank
 

输出：

silent

时间复杂度： O(N)，N 是不包括空格的字符串长度。
辅助空间： O(N)

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