Arrangement of words without changing the relative position of vowel and consonants?

Suppose we have a string with n elements (n < 10). We have to find the number of ways that the string can be arranged without changing the relative position of the vowel and consonants.

The approach is simple. We have to count the number of vowels and consonants in the given string, then we have to find how many ways we can arrange the vowels only, then find the number of ways to arrange the consonant only, after that multiply these two results to get the total ways.

Algorithm

arrangeWayCount(str)

Begin
   define an array ‘freq’ to store frequency.
   count and place frequency of each characters in freq array. such that freq[‘0’] will hold
   frequency of letter ‘a’, freq[1] will hold frequency of ‘b’ and so on.
   v := number of vowels, and c := number of consonants in str
   vArrange := factorial of v
   for each vowel v in [a, e, i, o, u], do
      vArrange := vArrange / factorial of the frequency of v
   done
   cArrange := factorial of c
   for each consonant con, do
      cArrange := cArrange / factorial of the frequency of con
   done
   return vArrange * cArrange
End

Example

#include <iostream>
using namespace std;
long long factorial(int n){
   if(n == 0 || n == 1)
      return 1;
   return n*factorial(n-1);
}
long long arrangeWayCount(string str){
   long long freq[27] = {0}; //fill frequency array to 0
   int v = 0, c = 0;
   for (int i = 0; i < str.length(); i++) {
      freq[str[i] - 'a']++;
      if (str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u') {
         v++;
      }else
         c++;
   }
   long long arrangeVowel;
   arrangeVowel = factorial(v);
   arrangeVowel /= factorial(freq[0]); // vowel a
   arrangeVowel /= factorial(freq[4]); // vowel e
   arrangeVowel /= factorial(freq[8]); // vowel i
   arrangeVowel /= factorial(freq[14]); // vowel o
   arrangeVowel /= factorial(freq[20]); // vowel u
   long long arrangeConsonant;
   arrangeConsonant = factorial(c);
   for (int i = 0; i < 26; i++) {
      if (i != 0 && i != 4 && i != 8 && i != 14 && i != 20)
      arrangeConsonant /= factorial(freq[i]); //frequency of all characters except vowels
   }
   long long total = arrangeVowel * arrangeConsonant;
   return total;
}
main() {
   string str = "computer";
   long long ans = arrangeWayCount(str);
   cout << "Possible ways to arrange: " << ans << endl;
}

Output

Possible ways to arrange: 720

Arnab Chakraborty

Published on 01-Aug-2019 07:40:38

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