Check whether all the substrings have number of vowels atleast as that of consonants

Namachila November 19, 2025

0 0 4 minutes read

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 vowel
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 sum
bool 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 code
int main()
{
    string str = "acaba";
    int n = str.length();
 
    if (isSatisfied(str, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function that returns 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;
    }
    return false;
}
 
// Compares two integers according
// to their digit sum
static 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 code
public 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 vowel
def isVowel(ch):
    if ch in ['i', 'a', 'e', 'o', 'u']:
        return True
    else:
        return False
 
# Compares two integers according
# to their digit sum
def 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 code
st = "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 approach
using System;
     
class GFG
{
 
// Function that returns 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;
    }
    return false;
}
 
// Compares two integers according
// to their digit sum
static 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 code
public 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.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!