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:
Input: S = “GADO”
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 <bits/stdc++.h>#define ll long long#define mod 1000000007#define N 1000001using namespace std;// Function to compute factorials till Nvoid Precomputefact(unordered_map<ll, ll>& 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 notbool 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 conditionvoid countAnagrams(string s, int n){ // Store the factorials upto N unordered_map<ll, ll> 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 Codeint main(){ string S = "GADO"; int l = S.size(); countAnagrams(S, l); return 0;} |
Python3
# Python 3 program for the above approach#include <bits/stdc++.h>mod = 1000000007N = 1000001fac = {}# Function to compute factorials till Ndef 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 notdef 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 conditiondef 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 Codeif __name__ == '__main__': S = "GADO" l = len(S) countAnagrams(S, l) # This code is contributed by ipg2016107. |
4
Time Complexity: O(N)
Auxiliary Space: O(N)
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