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++ 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 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);
}
} |
# 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# 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 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); |
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++ 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 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 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# 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. |
<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>
|
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++ 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;
} |
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);
} }
|
# Function to check if a character is a vowel or not def is_vowel(c):
return c.lower() in [ 'a' , 'e' , 'i' , 'o' , 'u' ]
# Function to check if the count of vowels in a string is even or not def is_equal_vowels(s):
count_vowels = sum ( 1 for char in s if is_vowel(char)) # Counting the number of vowels
# in the string
if count_vowels % 2 = = 0 : # If count of vowels is even, print 'Yes'
print ( "Yes" )
else : # If count of vowels is odd, print 'No'
print ( "No" )
def main():
s = "geeks" # Input string
is_equal_vowels(s)
if __name__ = = "__main__" :
main()
|
using System;
class Program
{ // Function to check if a 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' );
}
static void IsEqualVowels( string s)
{
int len = s.Length; // Length of the string
int countVowels = 0;
// Count the number of vowels present in the string
for ( int i = 0; i < len; i++)
{
if (IsVowel(s[i]))
countVowels++;
}
if (countVowels % 2 == 0) // If the count of vowels is even, then print "Yes"
Console.WriteLine( "Yes" );
else // Otherwise, print "No"
Console.WriteLine( "No" );
}
// Driver code static void Main()
{
string s = "geeks" ; // Input string
IsEqualVowels(s);
}
} |
// 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); |
Yes
Time Complexity: O(N), as traversing over the string.
Auxiliary Space: O(1)