Count of substrings consisting of even number of vowels

Given a string S of length N, the task is to find the number of non-empty substrings having even number of vowels.

Examples:

Input: N = 5, S = “abcde”
Output: 7
Explanation:
All possible substrings with even number of vowels are:
Substring Vowels
{abcde} 2
{b} 0
{bc} 0
{bcd} 0
{c} 0
{cd} 0
{d} 0
Input: N=4, S=”geeks”
Output: 6

Naive Approach:
The simplest approach to solve the problem is to generate all possible substrings of the given string and for each substring, count the number of vowels and check if it is even or not. If found to be even, increase count. Finally, after checking for all substrings, print the value of count as the answer.

Below is the implementation of the above approach:

C++

// C++ program to implement
//the above approach
#include <bits/stdc++.h>

using namespace std;
 
// Utility function to check 
// if a character is a vowel 

bool isVowel(char c)
{

    if (c == 'a' || c == 'e' || 

        c == 'i' || c == 'o' || 

        c == 'u')

        return true;
 
    return false;
}
 
// Function to calculate and return the
// count of substrings with even number
// of vowels

void countSubstrings(string s, int n)
{
 
    // Stores the count of substrings

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

    {

        int count = 0;

        for(int j = i; j < n; j++) 

        {
 
            // If the current character

            // is a vowel

            if (isVowel(s[j])) 

            {

                // Increase count

                count++;

            }
 
            // If substring contains

            // even number of vowels

            if (count % 2 == 0)
 
                // Increase the answer

                result++;

        }

    }
 
    // Print the final answer

    cout << result;
}
 
// Driver Code

int main()
{

    int n = 5;

    string s = "abcde";

    countSubstrings(s, n);

    return 0;
}
 
// This code is contributed by Amit Katiyar

Java

// Java Program to implement
// the above approach

class GFG {
 
    // Utility function to check

    // if a character is a vowel

    static boolean isVowel(char c)

    {

        if (c == 'a' || c == 'e' || c == 'i'

            || c == 'o' || c == 'u')

            return true;
 
        return false;

    }
 
    // Function to calculate and return the

    // count of substrings with even number

    // of vowels

    static void countSubstrings(String s, int n)

    {
 
        // Stores the count of substrings

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

            int count = 0;

            for (int j = i; j < n; j++) {
 
                // If the current character

                // is a vowel

                if (isVowel(s.charAt(j))) {
 
                    // Increase count

                    count++;

                }
 
                // If substring contains

                // even number of vowels

                if (count % 2 == 0)
 
                    // Increase the answer

                    result++;

            }

        }
 
        // Print the final answer

        System.out.println(result);

    }
 
    // Driver Code

    public static void main(String[] args)

    {

        int n = 5;

        String s = "abcde";

        countSubstrings(s, n);

    }
}

Python3

# Python3 Program to implement
# the above approach
 
# Utility function to check
# if a character is a vowel

def isVowel(c):
 
    if (c == 'a' or c == 'e' or

        c == 'i' or c == 'o' or

        c == 'u'):

        return True
 
    return False
 
# Function to calculate and return the
# count of substrings with even number
# of vowels

def countSubstrings(s, n):
 
    # Stores the count of substrings

    result = 0
 
    for i in range(n):

        count = 0

        for j in range(i, n):
 
            # If the current character

            # is a vowel

            if (isVowel(s[j])):
 
                #Increase count

                count += 1
 
            # If substring contains

            # even number of vowels

            if (count % 2 == 0):
 
                #Increase the answer

                result += 1
 
    # Print the final answer

    print(result)
 
# Driver Code

if __name__ == '__main__':
 
    n = 5

    s = "abcde"

    countSubstrings(s, n)
 
# This code is contributed by Mohit Kumar

// C# program to implement
// the above approach

using System;
 
class GFG{
 
// Utility function to check
// if a character is a vowel

static bool isVowel(char c)
{

    if (c == 'a' || c == 'e' || c == 'i' || 

        c == 'o' || c == 'u')

        return true;
 
    return false;
}
 
// Function to calculate and return the
// count of substrings with even number
// of vowels

static void countSubstrings(String s, int n)
{
 
    // Stores the count of substrings

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

    {

        int count = 0;

        for(int j = i; j < n; j++) 

        {
 
            // If the current character

            // is a vowel

            if (isVowel(s[j])) 

            {
 
                // Increase count

                count++;

            }
 
            // If substring contains

            // even number of vowels

            if (count % 2 == 0)
 
                // Increase the answer

                result++;

        }

    }
 
    // Print the final answer

    Console.WriteLine(result);
}
 
// Driver Code

public static void Main(String[] args)
{

    int n = 5;

    String s = "abcde";

    countSubstrings(s, n);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

<script>
 
    // Javascript program to implement

    // the above approach

    // Utility function to check 

    // if a character is a vowel 

    function isVowel(c)

    {

        if (c == 'a' || c == 'e' || 

            c == 'i' || c == 'o' || 

            c == 'u')

            return true;
 
        return false;

    }
 
    // Function to calculate and return the

    // count of substrings with even number

    // of vowels

    function countSubstrings(s, n)

    {
 
        // Stores the count of substrings

        let result = 0;
 
        for(let i = 0; i < n; i++)

        {

            let count = 0;

            for(let j = i; j < n; j++) 

            {
 
                // If the current character

                // is a vowel

                if (isVowel(s[j])) 

                {
 
                    // Increase count

                    count++;

                }
 
                // If substring contains

                // even number of vowels

                if (count % 2 == 0)
 
                    // Increase the answer

                    result++;

            }

        }
 
        // Print the final answer

        document.write(result);

    }
 
    let n = 5;

    let s = "abcde";

    countSubstrings(s, n);

</script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach:
Follow the steps below to optimize the above approach:

Initialize a cumulative count modulo array temp[], such that:

temp[0] : Stores count of substrings having even number of vowels.
temp[1] : Stores count of substrings having odd number of vowels.

Any substring [S_i, S_j] will contain even number of vowels if the number of vowels in [S₀, S_i-1], and the number of vowels in [S₀, S_j] have the same parity, i.e. either they are both even or both odd.
Since the count of substrings with even number of vowels and odd number of vowels are stored in temp[], then, by using handshaking lemma:

Total count of substrings = (temp[0] * (temp[0] – 1))/2 + (temp[1] * (temp[1] – 1))/2

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>

using namespace std;
 
// Utility function to check
// if a character is a vowel

bool isVowel(char c)
{

    if (c == 'a' || c == 'e' || 

        c == 'i' || c == 'o' || 

        c == 'u')

        return true;
 
    return false;
}
 
// Function to calculate and return the
// count of substrings with even number
// of vowels

void countSubstrings(string s, int n)
{

    // Stores the count of substrings

    // with even and odd number of

    // vowels respectively

    int temp[] = { 1, 0 };
 
    int result = 0, sum = 0;

    for(int i = 0; i <= n - 1; i++)

    {

        // Update count of vowels modulo 2

        // in sum to obtain even or odd

        sum += (isVowel(s[i]) ? 1 : 0);

        sum %= 2;
 
        // Increment even/odd count

        temp[sum]++;

    }
 
    // Count substrings with even number

    // of vowels using Handshaking Lemma

    result += ((temp[0] * (temp[0] - 1)) / 2);

    result += ((temp[1] * (temp[1] - 1)) / 2);
 
    cout << result;
}
 
// Driver Code

int main()
{

    int n = 5;

    string s = "abcde";

    countSubstrings(s, n);
}
 
// This code is contributed by Amit Katiyar

Java

// Java Program to implement
// the above approach

class GFG {
 
    // Utility function to check

    // if a character is a vowel

    static boolean isVowel(char c)

    {

        if (c == 'a' || c == 'e' || c == 'i'

            || c == 'o' || c == 'u')

            return true;
 
        return false;

    }
 
    // Function to calculate and return the

    // count of substrings with even number

    // of vowels

    static void countSubstrings(String s, int n)

    {

        // Stores the count of substrings

        // with even and odd number of

        // vowels respectively

        int temp[] = { 1, 0 };
 
        int result = 0, sum = 0;

        for (int i = 0; i <= n - 1; i++) {
 
            // Update count of vowels modulo 2

            // in sum to obtain even or odd

            sum += (isVowel(s.charAt(i)) ? 1 : 0);

            sum %= 2;
 
            // Increment even/odd count

            temp[sum]++;

        }
 
        // Count substrings with even number

        // of vowels using Handshaking Lemma

        result += ((temp[0] * (temp[0] - 1)) / 2);

        result += ((temp[1] * (temp[1] - 1)) / 2);
 
        System.out.println(result);

    }
 
    // Driver Code

    public static void main(String[] args)

    {

        int n = 5;

        String s = "abcde";

        countSubstrings(s, n);

    }
}

Python3

# Python3 program to implement
# the above approach
 
# Utility function to check
# if a character is a vowel

def isVowel(c):

    if (c == 'a' or c == 'e' or

        c == 'i' or c == 'o' or

        c == 'u'):

        return True;
 
    return False;
 
# Function to calculate and return the
# count of substrings with even number
# of vowels

def countSubstrings(s, n):

    # Stores the count of substrings

    # with even and odd number of

    # vowels respectively

    temp = [1, 0];
 
    result = 0;

    sum = 0;

    for i in range(0, n):

        # Update count of vowels modulo 2

        # in sum to obtain even or odd

        sum += (1 if isVowel(s[i]) else 0);

        sum %= 2;
 
        # Increment even/odd count

        temp[sum] += 1;
 
    # Count substrings with even number

    # of vowels using Handshaking Lemma

    result += ((temp[0] * (temp[0] - 1)) // 2);

    result += ((temp[1] * (temp[1] - 1)) // 2);
 
    print(result);
 
# Driver Code

if __name__ == '__main__':

    n = 5;

    s = "abcde";

    countSubstrings(s, n);
 
# This code is contributed by amal kumar choubey

// C# Program to implement
// the above approach

using System;

class GFG {

  // Utility function to check

  // if a character is a vowel

  static bool isVowel(char c)

  {

    if (c == 'a' || c == 'e' || c == 'i' || 

        c == 'o' || c == 'u')

      return true;
 
    return false;

  }
 
  // Function to calculate and return the

  // count of substrings with even number

  // of vowels

  static void countSubstrings(String s, int n)

  {

    // Stores the count of substrings

    // with even and odd number of

    // vowels respectively

    int []temp = { 1, 0 };
 
    int result = 0, sum = 0;

    for (int i = 0; i <= n - 1; i++) 

    {

      // Update count of vowels modulo 2

      // in sum to obtain even or odd

      sum += (isVowel(s[i]) ? 1 : 0);

      sum %= 2;
 
      // Increment even/odd count

      temp[sum]++;

    }
 
    // Count substrings with even number

    // of vowels using Handshaking Lemma

    result += ((temp[0] * (temp[0] - 1)) / 2);

    result += ((temp[1] * (temp[1] - 1)) / 2);
 
    Console.Write(result);

  }
 
  // Driver Code

  public static void Main(string[] args)

  {

    int n = 5;

    String s = "abcde";

    countSubstrings(s, n);

  }
}
 
// This code is contributed by rock_cool

Javascript

<script>

    // Javascript program to implement

    // the above approach

    // Utility function to check

    // if a character is a vowel

    function isVowel(c)

    {

        if (c == 'a' || c == 'e' ||

            c == 'i' || c == 'o' ||

            c == 'u')

            return true;
 
        return false;

    }
 
    // Function to calculate and return the

    // count of substrings with even number

    // of vowels

    function countSubstrings(s, n)

    {
 
        // Stores the count of substrings

        // with even and odd number of

        // vowels respectively

        let temp = [ 1, 0 ];
 
        let result = 0, sum = 0;

        for(let i = 0; i <= n - 1; i++)

        {
 
            // Update count of vowels modulo 2

            // in sum to obtain even or odd

            sum += (isVowel(s[i]) ? 1 : 0);

            sum %= 2;
 
            // Increment even/odd count

            temp[sum]++;

        }
 
        // Count substrings with even number

        // of vowels using Handshaking Lemma

        result += ((temp[0] * (temp[0] - 1)) / 2);

        result += ((temp[1] * (temp[1] - 1)) / 2);
 
        document.write(result);

    }

    let n = 5;

    let s = "abcde";

    countSubstrings(s, n);

    // This code is contributed by divyeshrabadiya07.
</script>

Output:

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

Article Tags :

DSA

Mathematical

Searching

Strings

substring

vowel-consonant