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Permutations of string such that no two vowels are adjacent

  • Difficulty Level : Expert
  • Last Updated : 26 May, 2021

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 t  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 = $7_{C_5} \times 5!$  .
Total no. of ways =  (6! / 2!) \times 7_{C_5} \times 5!
= 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 <bits/stdc++.h>
using namespace std;
 
// Factorial of a number
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)
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
int 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 code
int 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 number
def 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 adjacent
def 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 code
if __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




<?php
// CPP program to count permutations of string
// such that no two vowels are adjacent
  
// Factorial of a number
function factorial($n)
{
    $fact = 1;
    for ($i = 2; $i <= $n; $i++)
        $fact = $fact * $i;
  
    return $fact;
}
  
// Function to find c(n, r)
function ncr($n, $r)
{
    return factorial($n) / (factorial($r) * factorial($n - $r));
}
  
// Function to count permutations of string
// such that no two vowels are adjacent
function countWays($str)
{
    $freq = array_fill(0,26,NULL);
    $nvowels = 0;
    $nconsonants = 0;
  
    // Finding the frequencies of
    // the characters
    for ($i = 0; $i < strlen($str); $i++)
        ++$freq[ord($str[$i]) - ord('a')];
  
    // finding the no. of vowels and
    // consonants in given word
    for ($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 ($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 ($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




<script>
// Javascript program to count permutations of string
// such that no two vowels are adjacent
     
    // Factorial of a number
    function factorial(n)
    {
        let fact = 1;
            for (let i = 2; i <= n; i++)
                fact = fact * i;
           
            return fact;
    }
     
     // Function to find c(n, r)
    function ncr(n,r)
    {
        return Math.floor(factorial(n) / (factorial(r) * factorial(n - r)));
    }
     
    // Function to count permutations of string
        // such that no two vowels are adjacent
    function countWays(str)
    {
 
            let freq=new Array(26);
               
              for(let i=0;i<26;i++)
            {
                freq[i]=0;
            }
               
            let nvowels = 0, nconsonants = 0;
           
            let vplaces, cways, vways;
           
            // Finding the frequencies of
            // the characters
            for (let i = 0; i < str.length; i++)
                ++freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)];
           
            // finding the no. of vowels and
            // consonants in given word
            for (let 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 (let i = 0; i < 26; i++) {
                if (i != 0 && i != 4 && i != 8 && i != 14
                    && i != 20 && freq[i] > 1) {
           
                    cways = Math.floor(cways / factorial(freq[i]));
                }
            }
           
            // ways to put vowels 7C5 x 5!
            vways = ncr(vplaces, nvowels) * factorial(nvowels);
            for (let i = 0; i < 26; i++) {
                if (i == 0 || i == 4 || i == 8 || i == 14
                    || i == 20 && freq[i] > 1) {
                    vways = Math.floor(vways / factorial(freq[i]));
                }
            }
           
            return cways * vways;
             
    }
     
     // Driver code
    let str = "permutation";
    document.write(countWays(str));
     
     
     
// This code is contributed by rag2127
</script>

Output: 

907200

 

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