Count of adjacent Vowel Consonant Pairs

Given a string, the task is to count the number of adjacent pairs such that the first element of the pair is a consonant and the second element is a vowel. That is find the number of pairs (i, i+1) such that the ith character of this string is a consonant and the (i+1)th character is a vowel.

Examples:

Input :  str = "bazeci"
Output : 3

Input : str = "abu"
Output : 1

Algorithm:



  1. We have to find all possible adjacent consonant-vowel pairs.
  2. Insert all of the vowels in a set or hash, so that we can check if the current character is a vowel or consonant in constant time.
  3. We run a loop for the first n-1 elements and check, if the ith character is a consonant, and the (i+1)th character a vowel or not.
  4. If so, we increment the count, else we continue till the end of the string.

Below is the implementation of the above approach:

C++

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// C++ Program to implement the above approach
  
#include <bits/stdc++.h>
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<char> 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;
}

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Java

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// 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<Character> st=new HashSet<Character>(); 
    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

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Python3

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# 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

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C#

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// 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<char> st = new HashSet<char>(); 
    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

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Output:

3

Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1). We have used additional space to store vowels in a Hash but since number of vowels is only 5 so, the extra space used is considered as constant.



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