Check if a string can be split into two substrings with equal number of vowels

Given a string S, the task is to check if the string can be split into two substrings such that the number of vowels in both of them are equal. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: S = “geeks”
Output: Yes
Explanation: Splitting the strings into substrings “ge” (count of vowels = 1) and “eks” (count of vowels = 1) satisfies the condition.

Input: S = “textbook”
Output: No

Naive Approach: The simplest approach to solve this problem is to traverse the given string S and partition the string into two substrings at every possible index. For each such split, check if the two substrings have the same number of vowels present in them or not. If found to be true, then print “Yes” else print “No”. 

C++

// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if any
// character is a vowel or not
bool isVowel(char c)
{
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
            || c == 'u' || c == 'A' || c == 'E' || c == 'I'
            || c == 'O' || c == 'U');
}

// Function to check if string S
// can be split into two substrings
// with equal number of vowels
int containsEqualStrings(string S)
{
    int n = S.length();
    bool found = false;

    // Traversing in the given string
    for (int i = 0; i < n - 1; i++) {
        int vowelsLeft = 0, vowelsRight = 0;

        // Count the number of vowels in left substring
        for (int j = 0; j <= i; j++) {
            if (isVowel(S[j])) {
                vowelsLeft++;
            }
        }

        // Count the number of vowels in right substring
        for (int j = i + 1; j < n; j++) {
            if (isVowel(S[j])) {
                vowelsRight++;
            }
        }

        // check if substrings has equal
        // number of vowels
        if (vowelsLeft == vowelsRight) {
            found = true;
            break;
        }
    }

    if (found) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
}

// Driver Code
int main()
{
    string S = "geeks";
    containsEqualStrings(S);
}

// This code is contributed by Pushpesh Raj.

Java

// Java code for above approach
import java.io.*;
import java.util.*;

public class Main {

  // Function to check if any
  // character is a vowel or not
  static boolean isVowel(char c)
  {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
            || c == 'u' || c == 'A' || c == 'E' || c == 'I'
            || c == 'O' || c == 'U');
  }

  // Function to check if string S
  // can be split into two substrings
  // with equal number of vowels
  static int containsEqualStrings(String S)
  {
    int n = S.length();
    boolean found = false;

    // Traversing in the given string
    for (int i = 0; i < n - 1; i++) {
      int vowelsLeft = 0, vowelsRight = 0;

      // Count the number of vowels in left substring
      for (int j = 0; j <= i; j++) {
        if (isVowel(S.charAt(j))) {
          vowelsLeft++;
        }
      }

      // Count the number of vowels in right substring
      for (int j = i + 1; j < n; j++) {
        if (isVowel(S.charAt(j))) {
          vowelsRight++;
        }
      }

      // check if substrings has equal
      // number of vowels
      if (vowelsLeft == vowelsRight) {
        found = true;
        break;
      }
    }

    if (found) {
      System.out.println("Yes");
    }
    else {
      System.out.println("No");
    }
    return 0;
  }

  // Driver Code
  public static void main(String[] args)
  {
    String S = "geeks";
    containsEqualStrings(S);
  }
}

Python3

# Function to check if any
# character is a vowel or not
def isVowel(c):
    return (c == 'a' or c == 'e' or c == 'i' or c == 'o'
            or c == 'u' or c == 'A' or c == 'E' or c == 'I'
            or c == 'O' or c == 'U')

# Function to check if string S
# can be split into two substrings
# with equal number of vowels
def containsEqualStrings(str):
    n = len(str)
    found = False

    # Traversing in the given string
    for i in range(n - 1):
        vowelsLeft = 0
        vowelsRight = 0

        # Count the number of vowels in left substring
        for j in range(i + 1):
            if isVowel(str[j]):
                vowelsLeft += 1

        # Count the number of vowels in right substring
        for j in range(i + 1, n):
            if isVowel(str[j]):
                vowelsRight += 1

        # check if substrings has equal
        # number of vowels
        if vowelsLeft == vowelsRight:
            found = True
            break

    if found:
        print("Yes")
    else:
        print("No")

# Driver C
str = "geeks"
containsEqualStrings(str)

# This code is contributed by Jay

C#

// C# code for above approach
using System;

class Program {
    // Function to check if any
    // character is a vowel or not
    static bool IsVowel(char c)
    {
        return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
                || c == 'u' || c == 'A' || c == 'E'
                || c == 'I' || c == 'O' || c == 'U');
    }
    // Function to check if string S
    // can be split into two substrings
    // with equal number of vowels
    static void ContainsEqualStrings(string S)
    {
        int n = S.Length;
        bool found = false;

        // Traversing in the given string
        for (int i = 0; i < n - 1; i++) {
            int vowelsLeft = 0, vowelsRight = 0;

            // Count the number of vowels in left substring
            for (int j = 0; j <= i; j++) {
                if (IsVowel(S[j])) {
                    vowelsLeft++;
                }
            }

            // Count the number of vowels in right substring
            for (int j = i + 1; j < n; j++) {
                if (IsVowel(S[j])) {
                    vowelsRight++;
                }
            }

            // check if substrings has equal
            // number of vowels
            if (vowelsLeft == vowelsRight) {
                found = true;
                break;
            }
        }

        if (found) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }

    // Driver Code
    static void Main()
    {
        string S = "geeks";
        ContainsEqualStrings(S);
    }
}
// This code is contributed by sarojmcy2e

Javascript

// JavaScript code for above approach

// Function to check if any
// character is a vowel or not
function isVowel(c) {
    return (c === 'a' || c === 'e' || c === 'i' || c === 'o' ||
    c === 'u' || c === 'A' || c === 'E' || c === 'I' ||
    c === 'O' || c === 'U');
}

// Function to check if string S
// can be split into two substrings
// with equal number of vowels
function containsEqualStrings(S) {
    const n = S.length;
    let found = false;
    // Traversing in the given string
    for (let i = 0; i < n - 1; i++) {
        let vowelsLeft = 0, vowelsRight = 0;

        // Count the number of vowels in left substring
        for (let j = 0; j <= i; j++) {
            if (isVowel(S[j])) {
                vowelsLeft++;
            }
        }

        // Count the number of vowels in right substring
        for (let j = i + 1; j < n; j++) {
            if (isVowel(S[j])) {
                vowelsRight++;
            }
        }

        // check if substrings has equal
        // number of vowels
        if (vowelsLeft === vowelsRight) {
            found = true;
            break;
        }
    }

    if (found) {
        console.log("Yes");
    }
    else {
        console.log("No");
    }
}
// Driver Code
const S = "geeks";
containsEqualStrings(S);

Output

Yes

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

Efficient Approach: The above approach can be optimized by pre-calculating the total number of vowels in the string. Follow the steps below to solve the problem:

• Initialize two variables, say totalVowels and vowelsTillNow, to store the total number of vowels and the current count of vowels respectively.
• Now traverse the string S and count all the vowels and store it in the totalVowels.
• Now, again traverse the string S and decrement totalVowels by 1 and increment vowelsTillNow by 1, if a vowel occurs. Check if totalVowels become equal to vowelsTillNow at any point or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

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

// Function to check if any
// character is a vowel or not
bool isVowel(char ch)
{

    // Lowercase vowels
    if (ch == 'a' || ch == 'e' || ch == 'i'
        || ch == 'o' || ch == 'u')
        return true;

    // Uppercase vowels
    if (ch == 'A' || ch == 'E' || ch == 'I'
        || ch == 'O' || ch == 'U')
        return true;

    // Otherwise
    return false;
}

// Function to check if string S
// can be split into two substrings
// with equal number of vowels
string containsEqualStrings(string S)
{

    // Stores the count of vowels
    // in the string S
    int totalVowels = 0;

    // Traverse over the string
    for (int i = 0;
         i < S.size(); i++)
    {

        // If S[i] is vowel
        if (isVowel(S[i]))
            totalVowels++;
    }

    // Stores the count of vowels
    // upto the current index
    int vowelsTillNow = 0;

    // Traverse over the string
    for (int i = 0;
         i < S.size(); i++)
    {

        // If S[i] is vowel
        if (isVowel(S[i]))
        {
            vowelsTillNow++;
            totalVowels--;

            // If vowelsTillNow and
            // totalVowels are equal
            if (vowelsTillNow
                == totalVowels)
            {

                return "Yes";
            }
        }
    }

    // Otherwise
    return "No";
}

// Driver Code
int main()
{
    string S = "geeks";
    cout<<(containsEqualStrings(S));
}

// This code is contributed by mohit kumar 29.

Java

// Java program for the above approach

import java.io.*;

class GFG {

    // Function to check if any
    // character is a vowel or not
    public static boolean isVowel(char ch)
    {
        // Lowercase vowels
        if (ch == 'a' || ch == 'e' || ch == 'i'
            || ch == 'o' || ch == 'u')
            return true;

        // Uppercase vowels
        if (ch == 'A' || ch == 'E' || ch == 'I'
            || ch == 'O' || ch == 'U')
            return true;

        // Otherwise
        return false;
    }

    // Function to check if string S
    // can be split into two substrings
    // with equal number of vowels
    public static String
    containsEqualStrings(String S)
    {

        // Stores the count of vowels
        // in the string S
        int totalVowels = 0;

        // Traverse over the string
        for (int i = 0;
             i < S.length(); i++) {

            // If S[i] is vowel
            if (isVowel(S.charAt(i)))
                totalVowels++;
        }

        // Stores the count of vowels
        // upto the current index
        int vowelsTillNow = 0;

        // Traverse over the string
        for (int i = 0;
             i < S.length(); i++) {

            // If S[i] is vowel
            if (isVowel(S.charAt(i))) {
                vowelsTillNow++;
                totalVowels--;

                // If vowelsTillNow and
                // totalVowels are equal
                if (vowelsTillNow
                    == totalVowels) {

                    return "Yes";
                }
            }
        }

        // Otherwise
        return "No";
    }

    // Driver Code
    public static void main(String[] args)
    {
        String S = "geeks";
        System.out.println(
            containsEqualStrings(S));
    }
}

Python3

# Python3 program for the above approach

# Function to check if any
# character is a vowel or not
def isVowel(ch):

    # Lowercase vowels
    if (ch == 'a' or ch == 'e' or
        ch == 'i' or ch == 'o' or
        ch == 'u'):
        return True

    # Uppercase vowels
    if (ch == 'A' or ch == 'E' or
        ch == 'I' or ch == 'O' or
        ch == 'U'):
        return True

    # Otherwise
    return False

# Function to check if string S
# can be split into two substrings
# with equal number of vowels
def containsEqualStrings(S):

    # Stores the count of vowels
    # in the string S
    totalVowels = 0

    # Traverse over the string
    for i in range(len(S)):

        # If S[i] is vowel
        if (isVowel(S[i])):
            totalVowels += 1

    # Stores the count of vowels
    # upto the current index
    vowelsTillNow = 0

    # Traverse over the string
    for i in range(len(S)):

        # If S[i] is vowel
        if (isVowel(S[i])):
            vowelsTillNow += 1
            totalVowels -= 1

            # If vowelsTillNow and
            # totalVowels are equal
            if (vowelsTillNow == totalVowels):
                return "Yes"

    # Otherwise
    return "No"

# Driver Code
if __name__ == "__main__":

    S = "geeks"
    print(containsEqualStrings(S))

# This code is contributed by ukasp

C#

// C# program for the above approach
using System;
class GFG
{

    // Function to check if any
    // character is a vowel or not
    public static bool isVowel(char ch)
    {

        // Lowercase vowels
        if (ch == 'a' || ch == 'e' || ch == 'i'
            || ch == 'o' || ch == 'u')
            return true;

        // Uppercase vowels
        if (ch == 'A' || ch == 'E' || ch == 'I'
            || ch == 'O' || ch == 'U')
            return true;

        // Otherwise
        return false;
    }

    // Function to check if string S
    // can be split into two substrings
    // with equal number of vowels
    public static String
    containsEqualStrings(string S)
    {

        // Stores the count of vowels
        // in the string S
        int totalVowels = 0;

        // Traverse over the string
        for (int i = 0;
             i < S.Length; i++)
        {

            // If S[i] is vowel
            if (isVowel(S[i]))
                totalVowels++;
        }

        // Stores the count of vowels
        // upto the current index
        int vowelsTillNow = 0;

        // Traverse over the string
        for (int i = 0;
             i < S.Length; i++) {

            // If S[i] is vowel
            if (isVowel(S[i])) {
                vowelsTillNow++;
                totalVowels--;

                // If vowelsTillNow and
                // totalVowels are equal
                if (vowelsTillNow
                    == totalVowels) {

                    return "Yes";
                }
            }
        }

        // Otherwise
        return "No";
    }

// Driver Code
static public void Main()
{
    string S = "geeks";
    Console.WriteLine(
        containsEqualStrings(S));
}
}

// This code is contributed by code_hunt.

Javascript

<script>
      // JavaScript program for the above approach
      // Function to check if any
      // character is a vowel or not
      function isVowel(ch) {
        // Lowercase vowels
        if (ch === "a" || ch === "e" || ch === "i" || ch === "o" || ch === "u")
          return true;

        // Uppercase vowels
        if (ch === "A" || ch === "E" || ch === "I" || ch === "O" || ch === "U")
          return true;

        // Otherwise
        return false;
      }

      // Function to check if string S
      // can be split into two substrings
      // with equal number of vowels
      function containsEqualStrings(S) {
        // Stores the count of vowels
        // in the string S
        var totalVowels = 0;

        // Traverse over the string
        for (var i = 0; i < S.length; i++) {
          // If S[i] is vowel
          if (isVowel(S[i])) totalVowels++;
        }

        // Stores the count of vowels
        // upto the current index
        var vowelsTillNow = 0;

        // Traverse over the string
        for (var i = 0; i < S.length; i++) {
          // If S[i] is vowel
          if (isVowel(S[i])) {
            vowelsTillNow++;
            totalVowels--;

            // If vowelsTillNow and
            // totalVowels are equal
            if (vowelsTillNow === totalVowels) {
              return "Yes";
            }
          }
        }

        // Otherwise
        return "No";
      }

      // Driver Code
      var S = "geeks";
      document.write(containsEqualStrings(S));
    </script>

Output:

Yes

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

Approach 3: Count the number of vowels if the count is even then it is possible that we can divide it into two sub strings having same vowels else not.

This is simple and efficient approach. The following steps to follow to solve the problem:

1. Define a function  “isVowel” that returns true if the input character is a vowel else false.

2. Define another function “isEqualVowels” that takes a string input “s” and counts the number of vowels present in the string.

3. If the count of vowels is even, print “Yes” ( It means we can divide vowels into two sub-strings). If the count of vowels is odd, print “No”.

4.In the main function, taken a string “s” and call the “isEqualVowels” function with the string as an argument.

Below is the implementation of the above approach:

C++

// C++ code for above approach
#include <iostream>
using namespace std;

// Function to character is vowel or not.
bool isVowel(char c)
{
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
            || c == 'u' || c == 'A' || c == 'E' || c == 'I'
            || c == 'O' || c == 'U');
}

void isEqualVowels(string s)
{
  int len = s.length(); // length of the string
  int count_vowels = 0;

  // counting no. of vowels present in the string
  for(int i = 0; i<len; i++)
  {
    if(isVowel(s[i]))
      count_vowels++;
  }
  if(count_vowels%2==0)  // count of vowels is even the print Yes
     cout<<"Yes"<<endl;
   else                      // else print No
     cout<<"No"<<endl;
}

int main() {

    string s = "geeks"; // Input taken
   isEqualVowels(s);
    return 0;
}

Java

import java.io.*;

class GFG {

static boolean isVowel(char c)
{
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
            || c == 'u' || c == 'A' || c == 'E' || c == 'I'
            || c == 'O' || c == 'U');
}

static void isEqualVowels(String s)
{
  int len = s.length(); // length of the string
  int count_vowels = 0;

  // counting no. of vowels present in the string
  for(int i = 0; i<len; i++)
  {
    if(isVowel(s.charAt(i)))
      count_vowels++;
  }
  if(count_vowels%2==0)  // count of vowels is even the print Yes
     System.out.println("Yes");
   else                      // else print No
      System.out.println("No");
}

  // Driver code
public static void main(String []args)
    {
    String s = "geeks"; // Input taken
    isEqualVowels(s);
}
 }

Javascript

// JavaScript code for above approach

// Function to character is vowel or not.
function isVowel( c)
{
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o'
            || c == 'u' || c == 'A' || c == 'E' || c == 'I'
            || c == 'O' || c == 'U');
}

function isEqualVowels( s)
{
  let len = s.length; // length of the string
  let count_vowels = 0;

  // counting no. of vowels present in the string
  for(let i = 0; i<len; i++)
  {
    if(isVowel(s[i]))
      count_vowels++;
  }
  if(count_vowels%2==0)  // count of vowels is even the print Yes
     document.write("Yes");
   else                      // else print No
     document.write("No");
}

let s = "geeks"; // Input taken
isEqualVowels(s);

Output

Yes

Time Complexity: O(N), as traversing over the string.
Auxiliary Space: O(1)