Open In App

Count of substrings containing exactly K vowels

Improve
Improve
Like Article
Like
Save
Share
Report

Given string str containing both uppercase and lowercase letters, and an integer K. The task is to find the count of substrings containing exactly K vowels (maybe repetitive).

Examples:

Input: str = “aeiou”, K = 2
Output: 4
Explanation: The substrings are “ae”, “ei”, “io”, “ou”.

Input: str = “TrueGeek”, K = 3
Output: 5
Explanation: All such possible substrings are:
“TrueGe”, “rueGe”, “ueGe”, “eGee”, “eGeek”.

 

Approach: To solution of this problem is based on greedy approach. Generate all substrings and for each substring check the count of vowels. Follow the steps mentioned below.

  • Generate all the substrings. For each substring do the following
    • Store the count of occurrences of vowels.
    • Check if a new character in the substring is a vowel or not.
    • If it is a vowel, increment count of vowels found
    • Now for each substring, if the count of vowels is K, increment the final count.
  • Return the final count as the required answer at the end.

Below is the implementation of the above code.

C++




// C++ code to implement above approach
#include <bits/stdc++.h>
using namespace std;
  
#define MAX 128
  
// Function to check whether
// a character is vowel or not
bool isVowel(char x)
{
    return (x == 'a' || x == 'e' || 
            x == 'i' || x == 'o' || x == 'u' 
            || x == 'A' || x == 'E' || 
            x == 'I' || x == 'O' || x == 'U');
}
  
// Function to find the count of
// substring with k vowels
int get(string str, int k)
{
  
    int n = str.length();
  
    // Stores the count of
    // substring with K vowels
    int ans = 0;
  
    // Consider all substrings 
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
        int count = 0;
  
        // Consider all substrings 
        // between [i, j]
        for (int j = i; j < n; j++) {
  
            // If this is a vowel, for this
            // substring, increment count.
            if (isVowel(str[j])) {
                count++;
            }
  
            // If vowel count becomes k,
            // then increment final count.
            if (count == k) {
                ans++;
            }
  
            if (count > k)
                break;
        }
    }
    return ans;
}
  
// Driver code
int main(void)
{
    string s = "aeiou";
    int K = 2;
    cout << get(s, K);
    return 0;
}


Java




// Java code to implement above approach
class GFG
{
  
  static final int MAX = 128;
  
  // Function to check whether
  // a character is vowel or not
  static boolean isVowel(char x)
  {
    return (x == 'a' || x == 'e' || 
            x == 'i' || x == 'o' || x == 'u' 
            || x == 'A' || x == 'E' || 
            x == 'I' || x == 'O' || x == 'U');
  }
  
  // Function to find the count of
  // subString with k vowels
  static int get(String str, int k)
  {
  
    int n = str.length();
  
    // Stores the count of
    // subString with K vowels
    int ans = 0;
  
    // Consider all subStrings 
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int count = 0;
  
      // Consider all subStrings 
      // between [i, j]
      for (int j = i; j < n; j++) {
  
        // If this is a vowel, for this
        // subString, increment count.
        if (isVowel(str.charAt(j))) {
          count++;
        }
  
        // If vowel count becomes k,
        // then increment final count.
        if (count == k) {
          ans++;
        }
  
        if (count > k)
          break;
      }
    }
    return ans;
  }
  
  // Driver code
  public static void main(String[] args)
  {
    String s = "aeiou";
    int K = 2;
    System.out.print(get(s, K));
  }
}
  
// This code is contributed by 29AjayKumar


Python3




# python code to implement above approach
MAX = 128
  
# Function to check whether
# a character is vowel or not
def isVowel(x):
  
    return (x == 'a' or x == 'e' or
            x == 'i' or x == 'o' or x == 'u'
            or x == 'A' or x == 'E' or
            x == 'I' or x == 'O' or x == 'U')
  
# Function to find the count of
# substring with k vowels
def get(str, k):
  
    n = len(str)
  
    # Stores the count of
    # substring with K vowels
    ans = 0
  
    # Consider all substrings
    # beginning with str[i]
    for i in range(0, n):
        count = 0
  
        # Consider all substrings
        # between [i, j]
        for j in range(i, n):
  
            # If this is a vowel, for this
            # substring, increment count.
            if (isVowel(str[j])):
                count += 1
  
            # If vowel count becomes k,
            # then increment final count.
            if (count == k):
                ans += 1
  
            if (count > k):
                break
  
    return ans
  
# Driver code
if __name__ == "__main__":
  
    s = "aeiou"
    K = 2
    print(get(s, K))
  
# This code is contributed by rakeshsahni


C#




// C# code to implement above approach
using System;
class GFG {
  
  // Function to check whether
  // a character is vowel or not
  static bool isVowel(char x)
  {
    return (x == 'a' || x == 'e' || x == 'i' || x == 'o'
            || x == 'u' || x == 'A' || x == 'E'
            || x == 'I' || x == 'O' || x == 'U');
  }
  
  // Function to find the count of
  // substring with k vowels
  static int get(string str, int k)
  {
  
    int n = str.Length;
  
    // Stores the count of
    // substring with K vowels
    int ans = 0;
  
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int count = 0;
  
      // Consider all substrings
      // between [i, j]
      for (int j = i; j < n; j++) {
  
        // If this is a vowel, for this
        // substring, increment count.
        if (isVowel(str[j])) {
          count++;
        }
  
        // If vowel count becomes k,
        // then increment final count.
        if (count == k) {
          ans++;
        }
  
        if (count > k)
          break;
      }
    }
    return ans;
  }
  
  // Driver code
  public static void Main()
  {
    string s = "aeiou";
    int K = 2;
    Console.WriteLine(get(s, K));
  }
}
  
// This code is contributed by ukasp.


Javascript




<script>
   // JavaScript code for the above approach
 
   let MAX = 128
 
   // Function to check whether
   // a character is vowel or not
   function isVowel(x) {
     return (x == 'a' || x == 'e' ||
       x == 'i' || x == 'o' || x == 'u'
       || x == 'A' || x == 'E' ||
       x == 'I' || x == 'O' || x == 'U');
   }
 
   // Function to find the count of
   // substring with k vowels
   function get(str, k) {
 
     let n = str.length;
 
     // Stores the count of
     // substring with K vowels
     let ans = 0;
 
     // Consider all substrings 
     // beginning with str[i]
     for (let i = 0; i < n; i++) {
       let count = 0;
 
       // Consider all substrings 
       // between [i, j]
       for (let j = i; j < n; j++) {
 
         // If this is a vowel, for this
         // substring, increment count.
         if (isVowel(str[j])) {
           count++;
         }
 
         // If vowel count becomes k,
         // then increment final count.
         if (count == k) {
           ans++;
         }
 
         if (count > k)
           break;
       }
     }
     return ans;
   }
 
   // Driver code
   let s = "aeiou";
   let K = 2;
   document.write(get(s, K));
 
 // This code is contributed by Potta Lokesh
 </script>


 
 

Output

4

 

Time Complexity: O(N2) where N is the length of the string.
Auxiliary Space: O(1)

 



Last Updated : 29 Dec, 2021
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads