Given a string str consisting of lowercase English alphabets, the task is to count the number of adjacent pairs of vowels.
Examples:
Input: str = “abaebio”
Output: 2
(a, e) and (i, o) are the only valid pairs.
Input: str = “aeoui”
Output: 4
Approach: Starting from the first character of the string to the second last character, increment count for every character where str[i] as well as str[i + 1] are both vowels. Print the count in the end which is the required count of pairs.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that return true// if character ch is a vowelbool isVowel(char ch){ switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; default: return false; }}// Function to return the count of adjacent// vowel pairs in the given stringint vowelPairs(string s, int n){ int cnt = 0; for (int i = 0; i < n - 1; i++) { // If current character and the // character after it are both vowels if (isVowel(s[i]) && isVowel(s[i + 1])) cnt++; } return cnt;}// Driver codeint main(){ string s = "abaebio"; int n = s.length(); cout << vowelPairs(s, n); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function that return true // if character ch is a vowel static boolean isVowel(char ch) { switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; default: return false; } } // Function to return the count of adjacent // vowel pairs in the given string static int vowelPairs(String s, int n) { int cnt = 0; for (int i = 0; i < n - 1; i++) { // If current character and the // character after it are both vowels if (isVowel(s.charAt(i)) && isVowel(s.charAt(i + 1))) cnt++; } return cnt; } // Driver code public static void main(String args[]) { String s = "abaebio"; int n = s.length(); System.out.print(vowelPairs(s, n)); }} |
Python3
# Python3 implementation of the approach# Function that return true# if character ch is a voweldef isVowel(ch): if ch in ['a', 'e', 'i', 'o', 'u']: return True else: return False# Function to return the count of adjacent# vowel pairs in the given stringdef vowelPairs(s, n): cnt = 0 for i in range(n - 1): # If current character and the # character after it are both vowels if (isVowel(s[i]) and isVowel(s[i + 1])): cnt += 1 return cnt# Driver codes = "abaebio"n = len(s)print(vowelPairs(s, n))# This code is contributed# by mohit kumar |
C#
// C# implementation of the approach using System;class GFG { // Function that return true // if character ch is a vowel static bool isVowel(char ch) { switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; default: return false; } } // Function to return the count of adjacent // vowel pairs in the given string static int vowelPairs(string s, int n) { int cnt = 0; for (int i = 0; i < n - 1; i++) { // If current character and the // character after it are both vowels if (isVowel(s[i]) && isVowel(s[i + 1])) cnt++; } return cnt; } // Driver code public static void Main() { string s = "abaebio"; int n = s.Length; Console.WriteLine(vowelPairs(s, n)); } } // This code is contributed by Ryuga |
PHP
<?php// PHP implementation of the approach// Function that return true// if character ch is a vowelfunction isVowel($ch){ if($ch == 'a' || $ch == 'e' || $ch == 'i' || $ch == 'o' || $ch == 'u') return true; return false;}// Function to return the count of adjacent// vowel pairs in the given stringfunction vowelPairs($s, $n){ $cnt = 0; for ($i = 0; $i < $n - 1; $i++) { // If current character and the // character after it are both vowels if (isVowel($s[$i]) && isVowel($s[$i + 1])) $cnt++; } return $cnt;}// Driver code$s = "abaebio";$n = strlen($s);echo vowelPairs($s, $n);// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the approach // Function that return true // if character ch is a vowel function isVowel(ch) { switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; default: return false; } } // Function to return the count of adjacent // vowel pairs in the given string function vowelPairs(s, n) { let cnt = 0; for (let i = 0; i < n - 1; i++) { // If current character and the // character after it are both vowels if (isVowel(s[i]) && isVowel(s[i + 1])) cnt++; } return cnt; } // Driver Code let s = "abaebio"; let n = s.length; document.write(vowelPairs(s, n)); </script> |
Output:
2
Time Complexity: O(n) where n is the length of the string
Auxiliary Space: O(1)
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