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 = “neveropen”
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 notbool 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 vowelsint 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 Codeint main(){ string S = "neveropen"; containsEqualStrings(S);}// This code is contributed by Pushpesh Raj. |
Java
// Java code for above approachimport 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 = "neveropen"; containsEqualStrings(S); }} |
Python3
# Function to check if any# character is a vowel or notdef 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 vowelsdef 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 Cstr = "neveropen"containsEqualStrings(str)# This code is contributed by Jay |
C#
// C# code for above approachusing 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 = "neveropen"; ContainsEqualStrings(S); }}// This code is contributed by sarojmcy2e |
Javascript
// JavaScript code for above approach// Function to check if any// character is a vowel or notfunction 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 vowelsfunction 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 Codeconst S = "neveropen";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 notbool 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 vowelsstring 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 Codeint main(){ string S = "neveropen"; cout<<(containsEqualStrings(S));}// This code is contributed by mohit kumar 29. |
Java
// Java program for the above approachimport 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 = "neveropen"; System.out.println( containsEqualStrings(S)); }} |
Python3
# Python3 program for the above approach# Function to check if any# character is a vowel or notdef 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 vowelsdef 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 Codeif __name__ == "__main__": S = "neveropen" print(containsEqualStrings(S))# This code is contributed by ukasp |
C#
// C# program for the above approachusing 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 Codestatic public void Main(){ string S = "neveropen"; 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 = "neveropen"; 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 = "neveropen"; // 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 codepublic static void main(String []args) { String s = "neveropen"; // Input taken isEqualVowels(s);} } |
C#
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 = "neveropen"; // Input string 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 = "neveropen"; // Input takenisEqualVowels(s); |
Output
Yes
Time Complexity: O(N), as traversing over the string.
Auxiliary Space: O(1)
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