Given a string str, the task is to check whether all the substrings of length ? 2 have the number of vowels at least as that of the number of consonants.
Examples:
Input: str = “acaba”
Output: No
The substring “cab” has 2 consonants and a single vowel.
Input: str = “aabaa”
Output: Yes
Approach: There are only two cases where the given condition is not satisfied:
- When there are two consecutive consonants as in this case a substring of size 2 can have 2 consonants and no vowels.
- When there is a vowel surrounded by two consonants, in this case a substring of length 3 is possible with 2 consonants and 1 vowels.
All the other cases will always satisfy the given conditions.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true// if character ch is a vowelbool isVowel(char ch){ switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; } return false;}// Compares two integers according// to their digit sumbool isSatisfied(string str, int n){ // Check if there are two // consecutive consonants for (int i = 1; i < n; i++) { if (!isVowel(str[i]) && !isVowel(str[i - 1])) { return false; } } // Check if there is any vowel // surrounded by two consonants for (int i = 1; i < n - 1; i++) { if (isVowel(str[i]) && !isVowel(str[i - 1]) && !isVowel(str[i + 1])) { return false; } } return true;}// Driver codeint main(){ string str = "acaba"; int n = str.length(); if (isSatisfied(str, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachclass GFG{// Function that returns true// if character ch is a vowelstatic boolean isVowel(char ch){ switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; } return false;}// Compares two integers according// to their digit sumstatic boolean isSatisfied(char[] str, int n){ // Check if there are two // consecutive consonants for (int i = 1; i < n; i++) { if (!isVowel(str[i]) && !isVowel(str[i - 1])) { return false; } } // Check if there is any vowel // surrounded by two consonants for (int i = 1; i < n - 1; i++) { if (isVowel(str[i]) && !isVowel(str[i - 1]) && !isVowel(str[i + 1])) { return false; } } return true;}// Driver codepublic static void main(String []args) { String str = "acaba"; int n = str.length(); if (isSatisfied(str.toCharArray(), n)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach# Function that returns true# if acter ch is a voweldef isVowel(ch): if ch in ['i', 'a', 'e', 'o', 'u']: return True else: return False# Compares two integers according# to their digit sumdef isSatisfied(st, n): # Check if there are two # consecutive consonants for i in range(1, n): if (isVowel(st[i]) == False and isVowel(st[i - 1]) == False): return False # Check if there is any vowel # surrounded by two consonants for i in range(1, n - 1): if (isVowel(st[i]) and isVowel(st[i - 1]) == False and isVowel(st[i + 1]) == False ): return False return True# Driver codest = "acaba"n = len(st)if (isSatisfied(st, n)): print("Yes")else: print("No")# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System; class GFG{// Function that returns true// if character ch is a vowelstatic bool isVowel(char ch){ switch (ch) { case 'a': case 'e': case 'i': case 'o': case 'u': return true; } return false;}// Compares two integers according// to their digit sumstatic bool isSatisfied(char[] str, int n){ // Check if there are two // consecutive consonants for (int i = 1; i < n; i++) { if (!isVowel(str[i]) && !isVowel(str[i - 1])) { return false; } } // Check if there is any vowel // surrounded by two consonants for (int i = 1; i < n - 1; i++) { if (isVowel(str[i]) && !isVowel(str[i - 1]) && !isVowel(str[i + 1])) { return false; } } return true;}// Driver codepublic static void Main(String []args) { String str = "acaba"; int n = str.Length; if (isSatisfied(str.ToCharArray(), n)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript implementation of the approach // Function that returns true // if character ch is a vowel function isVowel(ch) { switch (ch) { case "a": case "e": case "i": case "o": case "u": return true; } return false; } // Compares two integers according // to their digit sum function isSatisfied(str, n) { // Check if there are two // consecutive consonants for (var i = 1; i < n; i++) { if (!isVowel(str[i]) && !isVowel(str[i - 1])) { return false; } } // Check if there is any vowel // surrounded by two consonants for (var i = 1; i < n - 1; i++) { if (isVowel(str[i]) && !isVowel(str[i - 1]) && !isVowel(str[i + 1])) { return false; } } return true; } // Driver code var str = "acaba"; var n = str.length; if (isSatisfied(str.split(""), n)) document.write("Yes"); else document.write("No"); </script> |
Output:
No
Time Complexity : O( | str | ) ,where | str | is length of given string str.
Auxiliary Space : O(1) ,as we are not using any extra space.
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