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

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>

Output

16

Time Complexity: O(N)
Auxiliary Space: O(1)