给定字符串S ，任务是检查字符串可以拆分为两个子字符串，以使两个子字符串中的元音数量相等。如果发现是真的，则打印“是” 。否则，打印“否” 。
例子：
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
天真的方法：解决此问题的最简单方法是遍历给定的字符串S，并将该字符串在每个可能的索引处分为两个子字符串。对于每个此类拆分，请检查两个子字符串中是否存在相同数量的元音。如果发现是真的，则打印“是”，否则打印“否” 。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效的方法：可以通过预先计算字符串的元音总数来优化上述方法。请按照以下步骤解决问题：
- 初始化两个变量，例如totalVowels和vowelsTillNow ，分别存储元音的总数和当前的元音计数。
- 现在遍历字符串S并计算所有元音并将其存储在totalVowels中。
- 现在，再次通过1和增量vowelsTillNow由1遍历字符串S和递减totalVowels，如果元音occurs.Check如果totalVowels变得在任何点或不等于vowelsTillNow。如果发现是真的，则打印“是” 。否则，打印“否” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
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 ukaspC#
// 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.输出：
Yes时间复杂度： O(N)
辅助空间： O(1)