# Permutations of string such that no two vowels are adjacent

• Difficulty Level : Expert
• Last Updated : 25 Jun, 2022

Given a string consisting of vowels and consonants. The task is to find the number of ways in which the characters of the string can be arranged such that no two vowels are adjacent to each other.
Note: Given that No. of vowels <= No. of consonants.

Examples:

Input: str = "permutation"
Output : 907200

Input: str = "geeksforgeeks"
Output: 3175200 

Approach:
Consider the above example string “permutation”:

• First place all of the consonants in the alternate places like below:
-- p -- r -- m -- t -- t -- n --
• Number of ways to place consonants = 6! / 2!. as appears twice and should be considered once.
• Then place the vowels in the remaining positions. We have 7 remaining positions and 5 vowels to fill these 7 places.
Therefore, the number of ways to fill vowels = .
Total no. of ways =  = 907200

Suppose, in a string, the number of vowels is vowelCount and the number of consonants is consonantCount

Therefore,
Total ways = (consonantCount! / duplicateConsonant!) * C(consonantCount+1 , vowelCount) * (vowelCount! / duplicateVowel!)

Below is the implementation of the above approach:

## C++

 // CPP program to count permutations of string// such that no two vowels are adjacent #include using namespace std; // Factorial of a numberint factorial(int n){     int fact = 1;    for (int i = 2; i <= n; i++)        fact = fact * i;     return fact;} // Function to find c(n, r)int ncr(int n, int r){    return factorial(n) / (factorial(r) * factorial(n - r));} // Function to count permutations of string// such that no two vowels are adjacentint countWays(string str){    int freq[26] = { 0 };    int nvowels = 0, nconsonants = 0;     int vplaces, cways, vways;     // Finding the frequencies of    // the characters    for (int i = 0; i < str.length(); i++)        ++freq[str[i] - 'a'];     // finding the no. of vowels and    // consonants in given word    for (int i = 0; i < 26; i++) {         if (i == 0 || i == 4 || i == 8            || i == 14 || i == 20)            nvowels += freq[i];        else            nconsonants += freq[i];    }    // finding places for the vowels    vplaces = nconsonants + 1;     // ways to fill consonants 6! / 2!    cways = factorial(nconsonants);    for (int i = 0; i < 26; i++) {        if (i != 0 && i != 4 && i != 8 && i != 14            && i != 20 && freq[i] > 1) {             cways = cways / factorial(freq[i]);        }    }     // ways to put vowels 7C5 x 5!    vways = ncr(vplaces, nvowels) * factorial(nvowels);    for (int i = 0; i < 26; i++) {        if (i == 0 || i == 4 || i == 8 || i == 14            || i == 20 && freq[i] > 1) {            vways = vways / factorial(freq[i]);        }    }     return cways * vways;} // Driver codeint main(){    string str = "permutation";     cout << countWays(str) << endl;     return 0;}

## Java

 // Java program to count permutations of string// such that no two vowels are adjacent class GFG{                     // Factorial of a number        static int factorial(int n)        {                     int fact = 1;            for (int i = 2; i <= n; i++)                fact = fact * i;                     return fact;        }                 // Function to find c(n, r)        static int ncr(int n, int r)        {            return factorial(n) / (factorial(r) * factorial(n - r));        }                 // Function to count permutations of string        // such that no two vowels are adjacent        static int countWays(String str)        {            int freq[]=new int[26];                         for(int i=0;i<26;i++)            {                freq[i]=0;            }                         int nvowels = 0, nconsonants = 0;                     int vplaces, cways, vways;                     // Finding the frequencies of            // the characters            for (int i = 0; i < str.length(); i++)                ++freq[str.charAt(i) - 'a'];                     // finding the no. of vowels and            // consonants in given word            for (int i = 0; i < 26; i++) {                         if (i == 0 || i == 4 || i == 8                    || i == 14 || i == 20)                    nvowels += freq[i];                else                    nconsonants += freq[i];            }            // finding places for the vowels            vplaces = nconsonants + 1;                     // ways to fill consonants 6! / 2!            cways = factorial(nconsonants);            for (int i = 0; i < 26; i++) {                if (i != 0 && i != 4 && i != 8 && i != 14                    && i != 20 && freq[i] > 1) {                             cways = cways / factorial(freq[i]);                }            }                     // ways to put vowels 7C5 x 5!            vways = ncr(vplaces, nvowels) * factorial(nvowels);            for (int i = 0; i < 26; i++) {                if (i == 0 || i == 4 || i == 8 || i == 14                    || i == 20 && freq[i] > 1) {                    vways = vways / factorial(freq[i]);                }            }                     return cways * vways;        }                 // Driver code        public static void main(String []args)        {            String str = "permutation";                     System.out.println(countWays(str));                          }} // This code is contributed// by ihritik

## Python3

 # Python3 program to count permutations of# string such that no two vowels are adjacent # Factorial of a numberdef factorial(n) :     fact = 1;    for i in range(2, n + 1) :        fact = fact * i     return fact # Function to find c(n, r)def ncr(n, r) :         return factorial(n) // (factorial(r) *                            factorial(n - r)) # Function to count permutations of string# such that no two vowels are adjacentdef countWays(string) :     freq = [0] * 26    nvowels, nconsonants = 0, 0     # Finding the frequencies of    # the characters    for i in range(len(string)) :        freq[ord(string[i]) - ord('a')] += 1     # finding the no. of vowels and    # consonants in given word    for i in range(26) :         if (i == 0 or i == 4 or i == 8            or i == 14 or i == 20) :            nvowels += freq[i]        else :            nconsonants += freq[i]         # finding places for the vowels    vplaces = nconsonants + 1     # ways to fill consonants 6! / 2!    cways = factorial(nconsonants)    for i in range(26) :        if (i != 0 and i != 4 and i != 8 and            i != 14 and i != 20 and freq[i] > 1) :             cways = cways // factorial(freq[i])     # ways to put vowels 7C5 x 5!    vways = ncr(vplaces, nvowels) * factorial(nvowels)    for i in range(26) :        if (i == 0 or i == 4 or i == 8 or i == 14            or i == 20 and freq[i] > 1) :            vways = vways // factorial(freq[i])     return cways * vways; # Driver codeif __name__ == "__main__" :     string = "permutation"     print(countWays(string)) # This code is contributed by Ryuga

## C#

 // C# program to count permutations of string// such that no two vowels are adjacent using System;class GFG{                     // Factorial of a number        static int factorial(int n)        {                     int fact = 1;            for (int i = 2; i <= n; i++)                fact = fact * i;                     return fact;        }                 // Function to find c(n, r)        static int ncr(int n, int r)        {            return factorial(n) / (factorial(r) * factorial(n - r));        }                 // Function to count permutations of string        // such that no two vowels are adjacent        static int countWays(String str)        {            int []freq=new int[26];                         for(int i=0;i<26;i++)            {                freq[i]=0;            }                         int nvowels = 0, nconsonants = 0;                     int vplaces, cways, vways;                     // Finding the frequencies of            // the characters            for (int i = 0; i < str.Length; i++)                ++freq[str[i] - 'a'];                     // finding the no. of vowels and            // consonants in given word            for (int i = 0; i < 26; i++) {                         if (i == 0 || i == 4 || i == 8                    || i == 14 || i == 20)                    nvowels += freq[i];                else                    nconsonants += freq[i];            }            // finding places for the vowels            vplaces = nconsonants + 1;                     // ways to fill consonants 6! / 2!            cways = factorial(nconsonants);            for (int i = 0; i < 26; i++) {                if (i != 0 && i != 4 && i != 8 && i != 14                    && i != 20 && freq[i] > 1) {                             cways = cways / factorial(freq[i]);                }            }                     // ways to put vowels 7C5 x 5!            vways = ncr(vplaces, nvowels) * factorial(nvowels);            for (int i = 0; i < 26; i++) {                if (i == 0 || i == 4 || i == 8 || i == 14                    || i == 20 && freq[i] > 1) {                    vways = vways / factorial(freq[i]);                }            }                     return cways * vways;        }                 // Driver code        public static void Main()        {            String str = "permutation";                     Console.WriteLine(countWays(str));                          }} // This code is contributed// by ihritik

## PHP

  1) {              $cways = $cways / factorial($freq[$i]);        }    }      // ways to put vowels 7C5 x 5!    $vways = ncr($vplaces, $nvowels) * factorial($nvowels);    for ($i = 0; $i < 26; $i++) { if ($i == 0 || $i == 4 || $i == 8 || $i == 14 || $i == 20 && $freq[$i] > 1) {            $vways = $vways / factorial($freq[$i]);        }    }    return $cways * $vways;}  // Driver code $str = "permutation";echo countWays($str)."\n";return 0;// this code is contributed by Ita_c.?>

## Javascript

 

Output:

907200

Time Complexity: O(|str|)

Auxiliary Space: O(1)

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