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Count of adjacent Vowel Consonant Pairs
  • Last Updated : 02 Jun, 2021

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++




// 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;
}

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<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

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<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

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to count the adjacent pairs of
// consonant and vowels in the String
function countPairs(s)
{
    // Using a set to store the vowels so that
    // checking each character becomes easier
    let st=new Set();
    st.add('a');
    st.add('e');
    st.add('i');
    st.add('o');
    st.add('u');
 
    // Variable to store number of
    // consonant-vowel pairs
    let count = 0;
 
    let n = s.length;
 
    for (let 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.has(s[i]) && !st.has(s[i + 1]))
            count++;
    }
 
    return count;
}
 
 
// Driver Code
 
     let s = "geeksforgeeks";
 
    document.write( countPairs(s));
  
 // This code is contributed by sanjoy_62.
</script>
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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