# Count anagrams having first character as a consonant and no pair of consonants or vowels placed adjacently

• Last Updated : 24 Feb, 2021

Given a string S of length N, the task is to count the number of anagrams of S whose first character is a consonant and no pair of consonants or vowels are adjacent to each other.

Examples:

Output: 4
Explanation:
The anagrams of string S satisfying the given conditions are GADO, GODA, DOGA, DAGO.
Therefore, the total number of such anagrams is 4.

Input: S = “AABCY”
Output: 6
Explanation:
The anagrams of the string S satisfying the given conditions are BACAY, BAYAC, CABAY, CAYAB, YABAC, YACAB.
Therefore, the total number of such anagrams is 6.

Naive Approach: The simplest approach is to generate all possible anagrams of the given string and count those anagrams that satisfy the given condition. Finally, print the count obtained.
Time Complexity: O(N!*N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the following observations:

• Strings that have an equal number of consonants and vowels satisfy the given condition.
• Strings having one more consonant than vowel also satisfy the given condition.
• Apart from these two conditions, the count of possible anagrams will always be 0.
• Now, the problem can be solved by using a combinatorial formula. Consider there are C1, C2…, CN consonants and V1, V2, …, VN vowels in the string S and and \sum C denote the total number of consonants and vowels respectively, then the answer would be: where,
Ci is the count of ith consonant.
Vi is the count of ith vowel.

Follow the steps below to solve the problem:

• Initialize a variable, say answer, to store the total count of anagrams.
• Store the frequency of each character of the string S in a HashMap count.
• Store the number of vowels and consonants in S in variables V and C respectively.
• If the value of V is not equal to C or C is not equal to (V + 1), then print 0. Otherwise, performing the following steps:
• Initialize denominator as 1.
• Traverse the string S using the variable i and update the denominator as denominator*((count[S[i]])!).
• Initialize numerator to V!*C!, and update the value of answer as numerator/denominator.
• After completing the above steps, print the value of the answer as the result.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include    #define ll long long #define mod 1000000007 #define N 1000001 using namespace std;   // Function to compute factorials till N void Precomputefact(unordered_map& fac) {     ll ans = 1;       // Iterate in the range [1, N]     for (ll i = 1; i <= N; i++) {           // Update ans to ans*i         ans = (ans * i) % mod;           // Store the value of ans         // in fac[i]         fac[i] = ans;     }     return; }   // Function to check whether the // current character is a vowel or not bool isVowel(char a) {     if (a == 'A' || a == 'E' || a == 'I' || a == 'O'         || a == 'U')         return true;     else         return false; }   // Function to count the number of // anagrams of S satisfying the // given condition void countAnagrams(string s, int n) {     // Store the factorials upto N     unordered_map fac;       // Function Call to generate     // all factorials upto n     Precomputefact(fac);       // Create a hashmap to store     // frequencies of all characters     unordered_map<char, ll> count;       // Store the count of     // vowels and consonants     int vo = 0, co = 0;       // Iterate through all     // characters in the string     for (int i = 0; i < n; i++) {           // Update the frequency         // of current character         count[s[i]]++;           // Check if the character         // is vowel or consonant         if (isVowel(s[i]))             vo++;         else             co++;     }       // Check if ΣC==ΣV+1 or ΣC==ΣV     if ((co == vo + 1) || (co == vo)) {           // Store the denominator         ll deno = 1;           // Calculate the denominator         // of the expression         for (auto c : count) {               // Multiply denominator by factorial             // of counts of all letters             deno = (deno * fac[c.second]) % mod;         }           // Store the numerator         ll nume = fac[co] % mod;         nume = (nume * fac[vo]) % mod;           // Store the answer by dividing         // numerator by denominator         ll ans = nume / deno;           // Print the answer         cout << ans;     }       // Otherwise, print 0     else {         cout << 0;     } }   // Driver Code int main() {     string S = "GADO";     int l = S.size();     countAnagrams(S, l);       return 0; }

## Python3

 # Python 3 program for the above approach #include    mod = 1000000007 N = 1000001   fac = {}   # Function to compute factorials till N def Precomputefact():     global fac     ans = 1       # Iterate in the range [1, N]     for i in range(1,N+1,1):         # Update ans to ans*i         ans = (ans * i) % mod           # Store the value of ans         # in fac[i]         fac[i] = ans       return   # Function to check whether the # current character is a vowel or not def isVowel(a):     if (a == 'A' or a == 'E' or a == 'I' or a == 'O' or a == 'U'):         return True     else:         return False   # Function to count the number of # anagrams of S satisfying the # given condition def countAnagrams(s,n):     # Store the factorials upto N     global fac       # Function Call to generate     # all factorials upto n     Precomputefact()       # Create a hashmap to store     # frequencies of all characters     count = {}       # Store the count of     # vowels and consonants     vo = 0     co = 0       # Iterate through all     # characters in the string     for i in range(n):         # Update the frequency         # of current character         if s[i] in count:             count[s[i]] += 1         else:             count[s[i]] = 1           # Check if the character         # is vowel or consonant         if (isVowel(s[i])):             vo += 1         else:             co += 1       # Check if ΣC==ΣV+1 or ΣC==ΣV     if ((co == vo + 1) or (co == vo)):         # Store the denominator         deno = 1           # Calculate the denominator         # of the expression         for key,value in count.items():             # Multiply denominator by factorial             # of counts of all letters             deno = (deno * fac[value]) % mod           # Store the numerator         nume = fac[co] % mod         nume = (nume * fac[vo]) % mod           # Store the answer by dividing         # numerator by denominator         ans = nume // deno           # Print the answer         print(ans)       # Otherwise, print 0     else:         print(0)   # Driver Code if __name__ == '__main__':     S = "GADO"     l = len(S)     countAnagrams(S, l)           # This code is contributed by ipg2016107.

Output:

4

Time Complexity: O(N)
Auxiliary Space: O(N)

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