Count the number of vowels occurring in all the substrings of given string | Set 2

Last Updated : 28 Jan, 2022

Given a string str[] of length N of lowercase characters containing 0 or more vowels, the task is to find the count of vowels that occurred in all the substrings of the given string.

Examples:

Input: str = “abc”
Output: 3
The given string “abc” contains only one vowel = ‘a’
Substrings of “abc” are = {“a”, “b”, “c”, “ab”, “bc, “abc”}
Hence, the sum of occurrences of the vowel in these strings = 3.(‘a’ occurred 3 times)

Input: str = “daceh”
Output: 16

Prefix Sum Approach: The Naive Approach and the approach using prefix sum are discussed in the Set 1 of this article.

Efficient Approach: Suppose, given string is str, substring starts from position x, and ends at position y and if the vowel are at i-th position, where 0<=x<=i and i<=y<-N . for each vowel, it could be in the substring, that is (i+1) choices for x and (n-i) choices for y. So there are total (i+1)*(n-i) substrings containing str[i]. Take a constant “aeiou“. Follow the steps below to solve the problem:

Initialize the variable res as 0 to store the answer.
Iterate over the range [0, N) using the variable i and perform the following tasks:
- If str[i] is a vowel then add the value of (I+1)*(N-i) to the variable res to count the total number of vowels possible.
After performing the above steps, print the value of res as the answer.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of vowels
long long countVowels(string str)
{
 
    // Define the size of the string
    // and ans as zero
    long res = 0, N = str.size();
 
    // Start traversing and find if their
    // is any vowel or not
    for (int i = 0; i < N; ++i)
 
        // If there is vowel, then count
        // the numbers of vowels
        if (string("aeiou").find(str[i])
            != string::npos)
            res += (i + 1) * (N - i);
 
    return res;
}
 
// Driver Code
int main()
{
    string str = "daceh";
 
    // Call the function
    long long ans = countVowels(str);
 
    cout << ans << endl;
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
  // Function to count the number of vowels
  static long countVowels(String str)
  {
 
    // Define the size of the String
    // and ans as zero
    long res = 0, N = str.length();
 
    // Start traversing and find if their
    // is any vowel or not
    for (int i = 0; i < N; ++i)
 
      // If there is vowel, then count
      // the numbers of vowels
      if (new String("aeiou").contains(String.valueOf(str.charAt(i))))
        res += (i + 1) * (N - i);
 
    return res;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String str = "daceh";
 
    // Call the function
    long ans = countVowels(str);
    System.out.print(ans +"\n");
 
  }
}
 
// This code is contributed by shikhasingrajput.

Python3

# Python program for the above approach
 
# Function to count the number of vowels
def countVowels(str):
   
    # Define the size of the String
    # and ans as zero
    res = 0;
    N = len(str);
 
    # Start traversing and find if their
    # is any vowel or not
    for i in range(N):
 
        # If there is vowel, then count
        # the numbers of vowels
        if ((str[i]) in ("aeiou")):
            res += (i + 1) * (N - i);
    return res;
 
# Driver Code
if __name__ == '__main__':
 
    str = "daceh";
 
    # Call the function
    ans = countVowels(str);
    print(ans);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
class GFG{
 
  // Function to count the number of vowels
  static long countVowels(String str)
  {
 
    // Define the size of the String
    // and ans as zero
    long res = 0, N = str.Length;
 
    // Start traversing and find if their
    // is any vowel or not
    for (int i = 0; i < N; ++i)
 
      // If there is vowel, then count
      // the numbers of vowels
      if (new String("aeiou").Contains(str[i]))
        res += (i + 1) * (N - i);
 
    return res;
  }
 
  // Driver Code
  public static void Main()
  {
    String str = "daceh";
 
    // Call the function
    long ans = countVowels(str);
    Console.Write(ans +"\n");
 
  }
}
 
// This code is contributed by gfgking

Javascript

<script>
// javascript program for the above approach 
// Function to count the number of vowels
  function countVowels(str)
  {
 
    // Define the size of the String
    // and ans as zero
    var res = 0, N = str.length;
 
    // Start traversing and find if their
    // is any vowel or not
    for (var i = 0; i < N; ++i)
 
      // If there is vowel, then count
      // the numbers of vowels
      if ("aeiou".includes(str[i]))
        res += (i + 1) * (N - i);
 
    return res;
  }
 
// Driver Code
var str = "daceh";
 
// Call the function
var ans = countVowels(str);
document.write(ans);
 
// This code is contributed by shikhasingrajput 
</script>