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Number of ways to arrange a word such that all vowels occur together

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Given a word containing vowels and consonants. The task is to find that in how many ways the word can be arranged so that the vowels always come together. Given that the length of the word <10.

Examples: 

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

Approach: Since word contains vowels and consonants together. All vowels are needed to remain together then we will take all vowels as a single letter. 

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 
 

Algorithm:

Step 1: Start
Step 2: Create a static function named “fact” of int return type and take the input value of int type as input which returns the factorial of a number.
Step 3: Now, create another static function of int return type named “waysOfConsonanats taking two parameters as input integer say “size1” and an integer array say “freq”.
             a. inside it let’s calculate the factorial of a given value that is sez1.
             b. Start a loop and iterate it 26 times and if the current value is a vowel, then continue the iteration else divide its c  calculated factorial by the given frequency in the freq array
             c. return the final value
Step 4: Now, create another static function of int return type named “waysOfVowels taking two parameters as input integer say  size2” and an integer array say “freq”.  
             a. Calculate the factorial of “size2” inside of “waysOfVowels” and divide it by the sum of the factorials for each vowel’s frequency.
Step 5: Create a static method called “countWays” that returns the overall number of possible word arrangements and accepts the string parameter “str.”
            a. Construct the 26-element “freq” integer array with all of its elements set to 0.
            b. Based on the ASCII value of the characterless “a,” loop over the characters in the string and increase the value of the appropriate “freq” element.
            c. The string’s vowels and consonants should be counted.
            d. By calling “waysOfConsonants” and “waysOfVowels” with the right parameters and multiplying the results, you can get the overall number of possible ways. 
Step 6: Return the final value
Step 7: End

Below is the implementation of the above approach: 

C++




// C++ program to calculate the no. of ways
// to arrange the word having vowels together
#include <bits/stdc++.h>
#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




<script>
// 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
</script>


Output: 

226800

 

Time complexity: O(n) where n is the length of the string
Auxiliary space: O(1) 

Further Optimizations: We can pre-compute required factorial values to avoid re-computations.



Last Updated : 28 Feb, 2023
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