计数有第一个字符作为辅音的字谜，并且没有一对相邻的辅音或元音

给定长度为N的字符串S ，任务是计算第一个字符是辅音且没有一对辅音或元音彼此相邻的S的字谜数目。

例子：

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.

天真的方法：最简单的方法是生成给定字符串的所有可能的字谜，并对满足给定条件的那些字谜进行计数。最后，打印获得的计数。
时间复杂度： O(N！* N)
辅助空间： O(1)

高效的方法：还可以基于以下观察来优化上述方法：

• 辅音和元音数目相等的字符串满足给定条件。
• 辅音比元音多一个的弦也满足给定条件。
• 除了这两个条件外，可能的字谜的计数始终为0 。
• 现在，可以通过使用组合公式来解决问题。考虑字符串S和C中有C ₁ ，C ₂ …，C _N辅音和V ₁ ，V ₂ ，…，V _N元音$\sum C$和\ sum C $\sum V$分别表示辅音和元音的总数，则答案将是：

$\frac{\sum C! * \sum V!}{(\sum C_1! *\sum C_2! * ... *C_N!)*(\sum V_1! *\sum V_2! * ... *V_N!)}$

where, 
C_i is the count of i^th consonant.
_Vi is the count of i^th vowel.

请按照以下步骤解决问题：

• 初始化一个变量，例如answer ，以存储字谜的总数。
• 将字符串S的每个字符的频率存储在HashMap计数中。
• 将元音和辅音的数量分别存储在变量V和C中的S中。
• 如果V的值不等于C或C的值不等于(V +1) ，则打印0 。否则，请执行以下步骤：
  • 将分母初始化为1 。
  • 使用变量i遍历字符串S ，并将分母更新为分母*((count [S [i]])！) 。
  • 将分子初始化为V！* C！ ，并将answer的值更新为分子/分母。
• 完成上述步骤后，打印答案的值作为结果。

下面是上述方法的实现：

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 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.

输出：

4

时间复杂度： O(N)
辅助空间： O(N)