# 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 ` `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

 ``

Output

`16`

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

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