计算第一个字符为辅音且没有相邻的辅音或元音对的字谜

给定长度为N的字符串S ，任务是计算S的第一个字符是辅音并且没有一对辅音或元音彼此相邻的字谜的数量。

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.

朴素的方法：最简单的方法是生成给定字符串的所有可能的字谜，并计算满足给定条件的字谜。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：上述方法也可以基于以下观察进行优化：

• 具有相同数量的辅音和元音的字符串满足给定条件。
• 辅音比元音多一个的字符串也满足给定条件。
• 除了这两个条件之外，可能的字谜计数将始终为0 。
• 现在，这个问题可以通过使用组合公式来解决。考虑在字符串S 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，那么答案将是：

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

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，比如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 count 中。
• 将S中元音和辅音的数量分别存储在变量V和C 中。
• 如果V的值不等于C或C不等于(V + 1) ，则打印0 。否则，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denominator*((count[S[i]])!) 。
  • 将分子初始化为V!*C! ，并将answer的值更新为numerator/denominator 。
• 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)

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