排列单词以使所有元音一起出现的方式的数量

给定一个包含元音和辅音的单词。任务是发现可以以多种方式排列单词，以使元音始终在一起。给定单词的长度<10。

例子：

Input: str = "geek"
Output: 6
Ways such that both 'e' comes together are 6 
i.e. geek, gkee, kgee, eekg, eegk, keeg

Input: str = "corporation"
Output: 50400

方法：由于单词同时包含元音和辅音。需要所有元音保持在一起，然后我们将所有元音当作一个字母。

As, in the word ‘geeksforgeeks’, we can treat the vowels “eeoee” as one letter. 
Thus, we have gksfrgks (eeoee). 
This has 9 (8 + 1) letters of which g, k, s each occurs 2 times and the rest are different.
The number of ways arranging these letters = 9!/(2!)x(2!)x(2!) = 45360 ways
Now, 5 vowels in which ‘e’ occurs 4 times and ‘o’ occurs 1 time, can be arranged in 5! /4! = 5 ways.
Required number of ways = (45360 x 5) = 226800 
 

下面是上述方法的实现：

C++

// C++ program to calculate the no. of ways
// to arrange the word having vowels together
#include 
#define ll long long int
using namespace std;
 
// Factorial of a number
ll fact(int n)
{
    ll f = 1;
    for (int i = 2; i <= n; i++)
        f = f * i;
    return f;
}
 
// calculating ways for arranging consonants
ll waysOfConsonants(int size1, int freq[])
{
    ll ans = fact(size1);
    for (int i = 0; i < 26; i++) {
 
        // Ignore vowels
        if (i == 0 || i == 4 || i == 8 || i == 14 || i == 20)
            continue;
        else
            ans = ans / fact(freq[i]);
    }
 
    return ans;
}
 
// calculating ways for arranging vowels
ll waysOfVowels(int size2, int freq[])
{
    return fact(size2) / (fact(freq[0]) * fact(freq[4]) * fact(freq[8])
                    * fact(freq[14]) * fact(freq[20]));
}
 
// Function to count total no. of ways
ll countWays(string str)
{
 
    int freq[26] = { 0 };
    for (int i = 0; i < str.length(); i++)
        freq[str[i] - 'a']++;
 
    // Count vowels and consonant
    int vowel = 0, consonant = 0;
    for (int i = 0; i < str.length(); i++) {
 
        if (str[i] != 'a' && str[i] != 'e' && str[i] != 'i'
            && str[i] != 'o' && str[i] != 'u')
            consonant++;
        else
            vowel++;
    }
 
    // total no. of ways
    return waysOfConsonants(consonant+1, freq) *
           waysOfVowels(vowel, freq);
}
 
// Driver code
int main()
{
    string str = "geeksforgeeks";
 
    cout << countWays(str) << endl;
 
    return 0;
}

Java

// Java program to calculate the no. of
// ways to arrange the word having
// vowels together
import java.util.*;
  
class GFG{
  
// Factorial of a number
static int fact(int n)
{
    int f = 1;
    for(int i = 2; i <= n; i++)
        f = f * i;
          
    return f;
}
  
// Calculating ways for arranging consonants
static int waysOfConsonants(int size1,
                            int []freq)
{
    int ans = fact(size1);
    for(int i = 0; i < 26; i++)
    {
          
        // Ignore vowels
        if (i == 0 || i == 4 || i == 8 ||
            i == 14 || i == 20)
            continue;
        else
            ans = ans / fact(freq[i]);
    }
    return ans;
}
  
// Calculating ways for arranging vowels
static int waysOfVowels(int size2, int [] freq)
{
    return fact(size2) / (fact(freq[0]) *
          fact(freq[4]) * fact(freq[8]) *
         fact(freq[14]) * fact(freq[20]));
}
  
// Function to count total no. of ways
static int countWays(String str)
{
    int []freq = new int [200];
    for(int i = 0; i < 200; i++)
        freq[i] = 0;
          
    for(int i = 0; i < str.length(); i++)
        freq[str.charAt(i) - 'a']++;
          
    // Count vowels and consonant
    int vowel = 0, consonant = 0;
    for(int i = 0; i < str.length(); i++)
    {
        if (str.charAt(i) != 'a' && str.charAt(i) != 'e' &&
            str.charAt(i) != 'i' && str.charAt(i) != 'o' &&
            str.charAt(i) != 'u')
            consonant++;
        else
            vowel++;
    }
  
    // Total no. of ways
    return waysOfConsonants(consonant + 1, freq) *
           waysOfVowels(vowel, freq);
}
  
// Driver code
public static void main(String []args)
{
    String str = "geeksforgeeks";
  
    System.out.println(countWays(str));
}
}
 
// This code is contributed by rutvik_56

Python3

# Python3 program to calculate
# the no. of ways to arrange
# the word having vowels together
 
# Factorial of a number
def fact(n):
 
    f = 1
    for i in range(2, n + 1):
        f = f * i
    return f
 
# calculating ways for
# arranging consonants
def waysOfConsonants(size1, freq):
 
    ans = fact(size1)
    for i in range(26):
 
        # Ignore vowels
        if (i == 0 or i == 4 or
            i == 8 or i == 14 or
            i == 20):
            continue
        else:
            ans = ans // fact(freq[i])
 
    return ans
 
# calculating ways for
# arranging vowels
def waysOfVowels(size2, freq):
 
    return (fact(size2) // (fact(freq[0]) *
            fact(freq[4]) * fact(freq[8]) *
            fact(freq[14]) * fact(freq[20])))
 
# Function to count total no. of ways
def countWays(str1):
 
    freq = [0] * 26
    for i in range(len(str1)):
        freq[ord(str1[i]) -
             ord('a')] += 1
 
    # Count vowels and consonant
    vowel = 0
    consonant = 0
    for i in range(len(str1)):
 
        if (str1[i] != 'a' and str1[i] != 'e' and
            str1[i] != 'i' and str1[i] != 'o' and
            str1[i] != 'u'):
            consonant += 1
        else:
            vowel += 1
 
    # total no. of ways
    return (waysOfConsonants(consonant + 1, freq) *
            waysOfVowels(vowel, freq))
 
# Driver code
if __name__ == "__main__":
 
    str1 = "geeksforgeeks"
    print(countWays(str1))
 
# This code is contributed by Chitranayal

C#

// C# program to calculate the no. of
// ways to arrange the word having
// vowels together
using System.Collections.Generic;
using System;
 
class GFG{
 
// Factorial of a number
static int fact(int n)
{
    int f = 1;
    for(int i = 2; i <= n; i++)
        f = f * i;
         
    return f;
}
 
// Calculating ways for arranging consonants
static int waysOfConsonants(int size1,
                            int []freq)
{
    int ans = fact(size1);
    for(int i = 0; i < 26; i++)
    {
         
        // Ignore vowels
        if (i == 0 || i == 4 || i == 8 ||
            i == 14 || i == 20)
            continue;
        else
            ans = ans / fact(freq[i]);
    }
    return ans;
}
 
// Calculating ways for arranging vowels
static int waysOfVowels(int size2, int [] freq)
{
    return fact(size2) / (fact(freq[0]) *
          fact(freq[4]) * fact(freq[8]) *
         fact(freq[14]) * fact(freq[20]));
}
 
// Function to count total no. of ways
static int countWays(string str)
{
    int []freq = new int [200];
    for(int i = 0; i < 200; i++)
        freq[i] = 0;
         
    for(int i = 0; i < str.Length; i++)
        freq[str[i] - 'a']++;
         
    // Count vowels and consonant
    int vowel = 0, consonant = 0;
    for(int i = 0; i < str.Length; i++)
    {
        if (str[i] != 'a' && str[i] != 'e' &&
            str[i] != 'i' && str[i] != 'o' &&
            str[i] != 'u')
            consonant++;
        else
            vowel++;
    }
 
    // Total no. of ways
    return waysOfConsonants(consonant + 1, freq) *
           waysOfVowels(vowel, freq);
}
 
// Driver code
public static void Main()
{
    string str = "geeksforgeeks";
 
    Console.WriteLine(countWays(str));
}
}
 
// This code is contributed by Stream_Cipher

Javascript

// Javascript program to calculate the no. of
// ways to arrange the word having
// vowels together
     
// Factorial of a number
function fact(n)
{
       let f = 1;
    for(let i = 2; i <= n; i++)
        f = f * i;
           
    return f;
}
 
// Calculating ways for arranging consonants
function waysOfConsonants(size1,freq)
{
    let ans = fact(size1);
    for(let i = 0; i < 26; i++)
    {
           
        // Ignore vowels
        if (i == 0 || i == 4 || i == 8 ||
            i == 14 || i == 20)
            continue;
        else
            ans = Math.floor(ans / fact(freq[i]));
    }
    return ans;
}
 
// Calculating ways for arranging vowels
function waysOfVowels(size2,freq)
{
    return Math.floor(fact(size2) / (fact(freq[0]) *
          fact(freq[4]) * fact(freq[8]) *
         fact(freq[14]) * fact(freq[20])));
}
 
// Function to count total no. of ways
function countWays(str)
{
    let freq = new Array(200);
    for(let i = 0; i < 200; i++)
        freq[i] = 0;
           
    for(let i = 0; i < str.length; i++)
        freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
           
    // Count vowels and consonant
    let vowel = 0, consonant = 0;
    for(let i = 0; i < str.length; i++)
    {
        if (str[i] != 'a' && str[i] != 'e' &&
            str[i] != 'i' && str[i] != 'o' &&
            str[i] != 'u')
            consonant++;
        else
            vowel++;
    }
   
    // Total no. of ways
    return waysOfConsonants(consonant + 1, freq) *
           waysOfVowels(vowel, freq);
}
 
 
// Driver code
let str = "geeksforgeeks";
document.write(countWays(str));
     
     
    // This code is contributed by avanitrachhadiya2155

输出：

226800

进一步优化：我们可以预先计算所需的阶乘值，以避免重新计算。