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:
// 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; } |
226800
Further Optimizations : We can pre-compute required factorial values to avoid re-computations.
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