给定句子形式的字符串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：
需要进行以下观察以解决该程序：
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：
- 为每个字母分配一个不同的素数。
- 对于给定字符串中的每个单词，计算分配给该单词的字符的素数乘积，并将其存储在HashMap中。
- 计算分配给单词字符的质数的乘积，并将其存储在HashMap中。
- 在HashMap中找到频率最高的产品，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 29C#
// 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 输出：
silent时间复杂度： O(N)，N是字符串的长度，不包括空格。
辅助空间： O(N)