Count the number of vowels occurring in all the substrings of given string

Given a string 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 

Naive Approach: Given a string of length N, the number of substrings that can be formed=N(N+1)/2. A simple solution is for each substring, we count the occurrences of the vowels and add them to get the result. The time complexity of this approach is O(N³) which is not suitable for large values of N.

Efficient Approach: The idea is to use a prefix sum array-based technique where we store the occurrences of each character in all the substrings concatenated. 

• For the first character, 
  

no. of occurrences = no. of substrings starting with the first character = N.

• For each of the following characters, we store the 
  

no. of substrings starting with that character + the number of substrings formed by the previous characters containing this character – the number of substrings formed by the previous characters only.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Returns the total sum of
// occurrences of all vowels
int vowel_calc(string s)
{
    int n = s.length();
    vector<int> arr;

    for (int i = 0; i < n; i++) {

        if (i == 0)
            // No. of occurrences of 0th character
            // in all the substrings
            arr.push_back(n);

        else
            // No. of occurrences of the ith character
            // in all the substrings
            arr.push_back((n - i) + arr[i - 1] - i);
    }

    int sum = 0;
    for (int i = 0; i < n; i++)

        // Check if ith character is a vowel
        if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
            || s[i] == 'o' || s[i] == 'u')
            sum += arr[i];

    // Return the total sum
    // of occurrences of vowels
    return sum;
}

// Driver code
int main()
{
    string s = "daceh";
    cout << vowel_calc(s) << endl;

    return 0;
}

Java

// Java implementation of the above approach

import java.io.*;
import java.util.*;

public class Gfg {

    // Returns the total sum of
    // occurrences of all vowels
    static int vowel_calc(String s)
    {
        int n = s.length();
        int arr[] = new int[n];

        for (int i = 0; i < n; i++) {

            if (i == 0)
                // No. of occurrences of 0th character
                // in all the substrings
                arr[i] = n;

            else
                // No. of occurrences of ith character
                // in all the substrings
                arr[i] = (n - i) + arr[i - 1] - i;
        }

        int sum = 0;
        for (int i = 0; i < n; i++) {
            char ch = s.charAt(i);
            // Check if ith character is a vowel
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u')
                sum += arr[i];
        }

        // Return the total sum
        // of occurrences of vowels
        return sum;
    }

    // Driver Code
    public static void main(String args[])
    {
        String s = "daceh";
        System.out.println(vowel_calc(s));
    }
}

Python3

# Python 3 implementation of
# a more efficient approach.
# return sum of all occurrences of all vowels
def sumVowel(string):
    n = len(string)
    sum = 0
    string = string.lower()

    # iterate through every character in the string
    for i in range(0, n):
        s = string[i]

        # checks if the character is a vowel or not
        if (s=="a" or s == "e" or s == "i" or s == "o" or s == "u"):

            # uses below expression to calculate the count
                        # of all occurrences of character in substrings
                        # of the string
            sum += ((n - i) * (i + 1))

    # return the total sum of occurrence
    return sum

#driver code
if __name__ == '__main__':
    #input string here
    string = "abhay"
    #print returned sum
    print(sumVowel(string))

# This code is contributed by
# Abhay Subramanian K

C#

// C# implementation of the above approach


using System;

public class Gfg {

    // Returns the total sum of
    // occurrences of all vowels
    static int vowel_calc(string s)
    {
        int n = s.Length;
        int[] arr = new int[n];

        for (int i = 0; i < n; i++) {

            if (i == 0)
                // No. of occurrences of 0th character
                // in all the substrings
                arr[i] = n;

            else
                // No. of occurrences of ith character
                // in all the substrings
                arr[i] = (n - i) + arr[i - 1] - i;
        }

        int sum = 0;
        for (int i = 0; i < n; i++) {
            char ch = s[i];
            // Check if ith character is a vowel
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u')
                sum += arr[i];
        }

        // Return the total sum
        // of occurrences of vowels
        return sum;
    }

    // Driver Code
    public static void Main()
    {
        string s = "daceh";
        Console.Write(vowel_calc(s));
    }
}

PHP

<?php
// PHP implementation of the above approach

// Returns the total sum of
// occurrences of all vowels
function vowel_calc($s)
{
    $n = strlen($s);
    $arr = array();

    for ($i = 0; $i < $n; $i++)
    {
        if ($i == 0)

            // No. of occurrences of 0th character
            // in all the substrings
            $arr[$i] = $n;

        else

            // No. of occurrences of ith character
            // in all the substrings
            $arr[$i] = ($n - $i) + $arr[$i - 1] - $i;
        }

    $sum = 0;
    for ($i = 0; $i < $n; $i++)
    {

        // Check if ith character is a vowel
        if ($s[$i] == 'a' || $s[$i] == 'e' ||
            $s[$i] == 'i' || $s[$i] == 'o' ||
            $s[$i] == 'u')
            $sum += $arr[$i];
        }

        // Return the total sum
        // of occurrences of vowels
        return $sum;
    }

// Driver Code
$s = "daceh";
echo(vowel_calc($s));

// This code is contributed by Shivi_Aggarwal
?>

Javascript

<script>

// Javascript implementation of the above approach

// Returns the total sum of
// occurrences of all vowels
function vowel_calc(s)
{
    var n = s.length;
    var arr = [];

    for (var i = 0; i < n; i++) {

        if (i == 0)
            // No. of occurrences of 0th character
            // in all the substrings
            arr.push(n);

        else
            // No. of occurrences of the ith character
            // in all the substrings
            arr.push((n - i) + arr[i - 1] - i);
    }

    var sum = 0;
    for (var i = 0; i < n; i++)

        // Check if ith character is a vowel
        if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
            || s[i] == 'o' || s[i] == 'u')
            sum += arr[i];

    // Return the total sum
    // of occurrences of vowels
    return sum;
}

// Driver code
var s = "daceh";
document.write( vowel_calc(s));

// This code is contributed by importantly.
</script>

Output

16

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

Another efficient approach without additional space:
The idea is based on the number of times a particular character can occur across all the substrings. If we know a particular character occurs x times in all the substrings, then we can just check if it is a vowel, and add the count to the sum.
A particular character A[i] can be part of substrings in the following two ways:
1. Substrings starting with A[i].
2. Substrings that contain character A[i] but do not start with A[i]

Observation: 
Let’s discuss this with an example:
s = “abcdefg”

Occurrences of character ‘a’:

• Substrings starting with ‘a’: “a”, “ab”, “abc” … “abcdefg” => 7 Count.
• No Substrings has ‘a’ in the middle.
  Total occurrences for a = 7 = 7 * 1

Occurrences of character ‘b’:

• Substrings starting with ‘b’: “b”, “bc”, … “bcdefg” => 6 Count.
• Substrings containing ‘b’ within them: 
  ‘b’ can also exist within substrings starting from a. Their count will be (7 – 1) = 6
  e.g.: “ab”, “abc” … “abcdefg”
  Total occurrences for b = 6 + 6 = 6 * 2

Occurrences of character ‘c’:

• Substrings starting with ‘c’: “c”, “cd” … “cdefg” => 5 Count.
• Substrings containing ‘c’ within them: 
  ‘c’ can also exist within substrings starting from a. e.g. “abc”, “abcd” … “abcdefg” (except ‘a’ and ‘ab’) = (7 – 2) = 5
  ‘c’ can also exist within substrings starting from b. e.g. “bc”, “bcd” … “bcdefg” (except ‘b’) = (6 – 1) = 5
  Total occurrences for c = 5 + 5 + 5 = 5 * 3

Here, a pattern in the number of occurrences can be observed, in which the total occurrences of the character will be equal to the product of the number of characters on the right side with the number of characters on the left side (including the character itself). So, whenever a vowel is encountered we can add the product of the count of right and left characters in the answer.

Below is the implementation of the above approach:

Java

// JAVA Code to implement the above approach

import java.io.*;
import java.util.*;

public class Gfg {

    // Function to return the total sum of
    // occurrences of all vowels
    static int vowel_calc(String s)
    {
        int n = s.length();

        // Variable to store the answer
        int totalOccurrences = 0;

        for (int i = 0; i < n; i++) {

            // Count of total occurrences of
            // current character
            int totalOccurrenceForCharInAllSubstrings
                = (i + 1) * (n - i);

            char ch = s.charAt(i);

            // if the current character is a vowel
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u') {

                // Add the occurrences of current
                // character to the answer
                totalOccurrences
                    += totalOccurenceForCharInAllSubstrings;
            }
        }

        // Return the total sum
        // of occurrences of vowels
        return totalOccurrences;
    }

    // Driver Code
    public static void main(String args[])
    {
        String s = "abcde";

          // Function Call
        System.out.println(vowel_calc(s));
    }
}

Output

10

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