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

Given a word containing vowels and consonants. The task is to find that in how many different ways 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 consonant 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

Below is the implementation of the above approach:

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// CPP 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;
}

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Output:

226800

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



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