相邻元音辅音对的计数

给定一个字符串，任务是计算相邻对的数量，以使该对中的第一个元素为辅音，第二个元素为元音。找出对(i，i + 1)的对数，以使该字符串的第i个字符为辅音，第(i + 1)个字符为元音。

例子：

Input :  str = "bazeci"
Output : 3

Input : str = "abu"
Output : 1

算法：

1. 我们必须找到所有可能的相邻辅音元音对。
2. 将所有元音插入集合或哈希中，以便我们可以在恒定时间内检查当前字符是元音还是辅音。
3. 我们对前n-1个元素运行一个循环，并检查第i个字符是否是辅音，第(i + 1)个字符是元音。
4. 如果是这样，我们将增加计数，否则我们将继续到字符串的末尾。

下面是上述方法的实现：

C++

// C++ Program to implement the above approach
  
#include 
using namespace std;
  
// Function to count the adjacent pairs of 
// consonant and vowels in the string
int countPairs(string s)
{
    // Using a set to store the vowels so that
    // checking each character becomes easier
    set st;
    st.insert('a');
    st.insert('e');
    st.insert('i');
    st.insert('o');
    st.insert('u');
  
    // Variable to store number of
    // consonant-vowel pairs
    int count = 0;
  
    int n = s.size();
  
    for (int i = 0; i < n - 1; i++) {
  
        // If the ith character is not found in the set,
        // means it is a consonant
        // And if the (i+1)th character is found in the set,
        // means it is a vowel
        // We increment the count of such pairs
        if (st.find(s[i]) == st.end() && st.find(s[i + 1]) != st.end())
            count++;
    }
  
    return count;
}
  
// Driver Code
int main()
{
    string s = "geeksforgeeks";
  
    cout << countPairs(s);
  
    return 0;
}

Java

// Java Program to implement the above approach 
import java.util.*;
  
class Sol
{
      
// Function to count the adjacent pairs of 
// consonant and vowels in the String 
static int countPairs(String s) 
{ 
    // Using a set to store the vowels so that 
    // checking each character becomes easier 
    Set st=new HashSet(); 
    st.add('a'); 
    st.add('e'); 
    st.add('i'); 
    st.add('o'); 
    st.add('u'); 
  
    // Variable to store number of 
    // consonant-vowel pairs 
    int count = 0; 
  
    int n = s.length(); 
  
    for (int i = 0; i < n - 1; i++)
    { 
  
        // If the ith character is not found in the set, 
        // means it is a consonant 
        // And if the (i+1)th character is found in the set, 
        // means it is a vowel 
        // We increment the count of such pairs 
        if (st.contains(s.charAt(i)) && !st.contains(s.charAt(i + 1))) 
            count++; 
    } 
  
    return count; 
} 
  
// Driver Code 
public static void main(String args[])
{ 
    String s = "geeksforgeeks"; 
  
    System.out.println( countPairs(s)); 
}
}
  
// This code is contributed by Arnab Kundu

Python3

# Python3 Program to implement the above approach 
  
# Function to count the adjacent pairs of 
# consonant and vowels in the string 
def countPairs(s) : 
  
    # Using a set to store the vowels so that 
    # checking each character becomes easier 
    st = set(); 
    st.add('a'); 
    st.add('e'); 
    st.add('i'); 
    st.add('o'); 
    st.add('u'); 
  
    # Variable to store number of 
    # consonant-vowel pairs 
    count = 0; 
  
    n = len(s); 
  
    for i in range(n - 1) :
          
        # If the ith character is not found in the set, 
        # means it is a consonant 
        # And if the (i+1)th character is found in the set, 
        # means it is a vowel 
        # We increment the count of such pairs 
        if (s[i] not in st and s[i + 1] in st) : 
            count += 1; 
  
    return count; 
  
# Driver Code 
if __name__ == "__main__" : 
      
    s = "geeksforgeeks"; 
  
    print(countPairs(s)); 
      
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic; 
  
class GFG
{
      
// Function to count the adjacent pairs of 
// consonant and vowels in the String 
static int countPairs(String s) 
{ 
    // Using a set to store the vowels so that 
    // checking each character becomes easier 
    HashSet st = new HashSet(); 
    st.Add('a'); 
    st.Add('e'); 
    st.Add('i'); 
    st.Add('o'); 
    st.Add('u'); 
  
    // Variable to store number of 
    // consonant-vowel pairs 
    int count = 0; 
  
    int n = s.Length; 
  
    for (int i = 0; i < n - 1; i++)
    { 
  
        // If the ith character is not found in the set, 
        // means it is a consonant 
        // And if the (i+1)th character is found in the set, 
        // means it is a vowel 
        // We increment the count of such pairs 
        if (st.Contains(s[i]) && !st.Contains(s[i + 1])) 
            count++; 
    } 
  
    return count; 
} 
  
// Driver Code 
public static void Main(String[] args)
{ 
    String s = "geeksforgeeks"; 
  
    Console.Write( countPairs(s)); 
}
}
  
// This code has been contributed by 29AjayKumar

输出：

3

时间复杂度：O(N)，其中N是字符串的长度。
辅助空间：O(1)。我们已经使用了额外的空间来将元音存储在哈希中，但是由于元音的数量只有5个，因此所使用的额外空间被认为是恒定的。