仅由元音组成的子串计数

给定一个字符串S ，任务是计算所有只包含元音的子字符串。

例子：

Input: S = “geeksforgeeks” 
Output: 7 
Explanation: 
Substrings {“e”, “ee”, “e”, “o”, “e”, “ee”, “e”} consists only of vowels.

Input: S = “aecui” 
Output: 6 
Explanation: 
Substrings {“a”, “ae”, “e”, “u”, “ui”, “i”} consists only of vowels. 

天真的方法：
最简单的方法是生成所有子串并检查它们中的每一个是否只包含元音。

下面是上述方法的实现：

C++

// C++ program to Count all substrings
// in a string which contains only vowels
#include 
using namespace std;
 
// Function to check if a
// character is vowel or not
bool isvowel(char ch)
{
    return (ch == 'a' or ch == 'e'
            or ch == 'i' or ch == 'o'
            or ch == 'u');
}
 
// Function to check whether
// string contains only vowel
bool isvalid(string& s)
{
    int n = s.length();
 
    for (int i = 0; i < n; i++) {
        // Check if the character is
        // not vowel then invalid
        if (!isvowel(s[i]))
            return false;
    }
 
    return true;
}
 
// Function to Count all substrings
// in a string which contains
// only vowels
int CountTotal(string& s)
{
    int ans = 0;
 
    int n = s.length();
 
    // Generate all substring of s
    for (int i = 0; i < n; i++) {
        string temp = "";
        for (int j = i; j < n; j++) {
            temp += s[j];
 
            // If temp contains only vowels
            if (isvalid(temp))
                // Increment the count
                ans += 1;
        }
    }
 
    return ans;
}
 
// Driver Program
int main()
{
    string s = "aeoibsddaaeiouudb";
 
    cout << (CountTotal(s)) << endl;
 
    return 0;
}

Java

// Java program to count all subStrings
// in a String which contains only vowels
import java.util.*;
 
class GFG{
 
// Function to check if a
// character is vowel or not
static boolean isvowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// Function to check whether
// String contains only vowel
static boolean isvalid(String s)
{
    int n = s.length();
 
    for(int i = 0; i < n; i++)
    {
         
       // Check if the character is
       // not vowel then invalid
       if (!isvowel(s.charAt(i)))
           return false;
    }
    return true;
}
 
// Function to Count all subStrings
// in a String which contains
// only vowels
static int CountTotal(String s)
{
    int ans = 0;
    int n = s.length();
 
    // Generate all subString of s
    for(int i = 0; i < n; i++)
    {
       String temp = "";
       for(int j = i; j < n; j++)
       {
          temp += s.charAt(j);
           
          // If temp contains only vowels
          if (isvalid(temp))
               
              // Increment the count
              ans += 1;
       }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "aeoibsddaaeiouudb";
 
    System.out.print((CountTotal(s)) + "\n");
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to count all substrings
# in a string which contains only vowels
 
# Function to check if a
# character is vowel or not
def isvowel(ch):
    return (ch == 'a' or ch == 'e' or
            ch == 'i' or ch == 'o' or
            ch == 'u')
 
# Function to check whether
# string contains only vowel
def isvalid(s):
     
    n = len(s)
    for i in range(n):
         
        # Check if the character is
        # not vowel then invalid
        if (not isvowel(s[i])):
            return False
             
    return True
 
# Function to Count all substrings
# in a string which contains
# only vowels
def CountTotal(s):
     
    ans = 0
    n = len(s)
 
    # Generate all substring of s
    for i in range(n):
        temp = ""
         
        for j in range(i, n):
            temp += s[j]
 
            # If temp contains only vowels
            if (isvalid(temp)):
                 
                # Increment the count
                ans += 1
                 
    return ans
 
# Driver code
if __name__ == '__main__':
     
    s = "aeoibsddaaeiouudb"
 
    print(CountTotal(s))
 
# This code is contributed by mohit kumar 29

C#

// C# program to count all subStrings
// in a String which contains only vowels
using System;
class GFG{
 
// Function to check if a
// character is vowel or not
static Boolean isvowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// Function to check whether
// String contains only vowel
static Boolean isvalid(string s)
{
    int n = s.Length;
 
    for(int i = 0; i < n; i++)
    {
         
        // Check if the character is
        // not vowel then invalid
        if (!isvowel(s[i]))
            return false;
    }
    return true;
}
 
// Function to Count all subStrings
// in a String which contains
// only vowels
static int CountTotal(string s)
{
    int ans = 0;
    int n = s.Length;
 
    // Generate all subString of s
    for(int i = 0; i < n; i++)
    {
        string temp = "";
        for(int j = i; j < n; j++)
        {
            temp += s[j];
             
            // If temp contains only vowels
            if (isvalid(temp))
                 
                // Increment the count
                ans += 1;
        }
    }
    return ans;
}
 
// Driver code
public static void Main()
{
    string s = "aeoibsddaaeiouudb";
 
    Console.Write((CountTotal(s)) + "\n");
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript program to count all subStrings
// in a String which contains only vowels
 
// Function to check if a
// character is vowel or not
function isvowel(ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// Function to check whether
// String contains only vowel
function isvalid(s)
{
    let n = s.length;
 
    for(let i = 0; i < n; i++)
    {
 
        // Check if the character is
        // not vowel then invalid
        if (!isvowel(s[i]))
            return false;
    }
    return true;
}
 
// Function to Count all subStrings
// in a String which contains
// only vowels
function CountTotal(s)
{
    let ans = 0;
    let n = s.length;
 
    // Generate all subString of s
    for(let i = 0; i < n; i++)
    {
        let temp = "";
        for(let j = i; j < n; j++)
        {
            temp += s[j];
 
            // If temp contains only vowels
            if (isvalid(temp))
 
                // Increment the count
                ans += 1;
        }
    }
    return ans;
}
 
// Driver code
let s = "aeoibsddaaeiouudb";
 
document.write((CountTotal(s)));
 
// This code is contributed by divyesh072019
 

输出：

38

时间复杂度： O(N ³ )

有效的方法：
为了优化上述方法，主要思想是计算只包含元音的子串的长度，比如 x。然后对于每个x ，可能的子串的数量是x * (x + 1) / 2只包含元音。对每个这样的子字符串重复此过程并返回最终计数。

下面是上述方法的实现：

C++

// C++ program to Count all substrings
// in a string which contains only vowels
#include 
using namespace std;
 
// Function to check vowel or not
bool isvowel(char ch)
{
    return (ch == 'a' or ch == 'e'
            or ch == 'i' or ch == 'o'
            or ch == 'u');
}
 
// Function to Count all substrings
// in a string which contains
// only vowels
int CountTotal(string& s)
{
    int ans = 0;
    int n = s.length();
    int cnt = 0;
 
    for (int i = 0; i < n; i++) {
        // Check if current character
        // is vowel
        if (isvowel(s[i]))
            // Increment length of substring
            cnt += 1;
 
        else {
            // Calculate possible
            // substrings of calculated
            // length
            ans += (cnt * (cnt + 1) / 2);
 
            // Reset the length
            cnt = 0;
        }
    }
 
    // Add remaining possible
    // substrings consisting
    // of vowels occupying
    // last indices of the string
    if (cnt != 0) {
        ans += (cnt * (cnt + 1) / 2);
    }
 
    return ans;
}
 
// Driver Program
int main()
{
    string s = "geeksforgeeks";
 
    cout << (CountTotal(s)) << endl;
 
    return 0;
}

Java

// Java program to Count all substrings
// in a string which contains only vowels
import java.io.*;
 
public class GFG {
 
    // Function to check vowel or not
    static boolean isvowel(char x)
    {
        return (x == 'a' || x == 'e'
                || x == 'i' || x == 'o'
                || x == 'u');
    }
 
    // Function to find largest
    // string which satisy condition
    static int CountTotal(String str)
    {
        int ans = 0;
 
        int n = str.length();
 
        char[] s = str.toCharArray();
 
        int cnt = 0;
 
        for (int i = 0; i < n; i++) {
            // Check if current character
            // is vowel
            if (isvowel(s[i]))
                // Increment count
                cnt += 1;
 
            else {
                // Count all possible
                // substrings of calculated
                // length
                ans += (cnt * (cnt + 1) / 2);
 
                // Reset the length
                cnt = 0;
            }
        }
 
        // Add remaining possible
        // substrings consisting
        // of vowels occupying
        // last indices of the string
        if (cnt != 0)
 
            ans += (cnt * (cnt + 1) / 2);
 
        return ans;
    }
 
    // Driver Program
    static public void main(String[] args)
    {
        String s = "geeksforgeeks";
 
        System.out.println(CountTotal(s));
    }
}

蟒蛇3

# Python3 program to Count all substrings
# in a string which contains only vowels
 
# function to check vowel or not
def isvowel(ch):
    return(ch in "aeiou")
 
 
# Function to Count all substrings
# in a string which contains
# only vowels
def CountTotal(s):
     
    ans = 0
    n = len(s)
    cnt = 0
     
    for i in range(0, n):
         
        # if current character is
        # vowel
        if(isvowel(s[i])):
            # increment
            cnt += 1
             
                 
        else:
             
            # Count all possible
            # substring of calculated
            # length
            ans += (cnt*(cnt + 1) // 2)
                 
            # Reset the length
            cnt = 0
 
    # Add remaining possible
    # substrings consisting
    # of vowels occupying
    # last indices of the string
    if(cnt != 0):
        ans += (cnt*(cnt + 1) // 2)
 
    return ans
 
# Driver Program
s = "geeksforgeeks"
 
print(CountTotal(s))

C＃

// C# program to count all substrings
// in a string which contains only vowels
using System;
 
class GFG{
 
// Function to check vowel or not
static bool isvowel(char x)
{
    return (x == 'a' || x == 'e' ||
            x == 'i' || x == 'o' ||
            x == 'u');
}
 
// Function to find largest
// string which satisy condition
static int CountTotal(string str)
{
    int ans = 0;
    int n = str.Length;
    char[] s = str.ToCharArray();
    int cnt = 0;
 
    for(int i = 0; i < n; i++)
    {
        
       // Check if current character
       // is vowel
       if (isvowel(s[i]))
        
           // Increment count
           cnt += 1;
       else
       {
            
           // Count all possible
           // substrings of calculated
           // length
           ans += (cnt * (cnt + 1) / 2);
            
           // Reset the length
           cnt = 0;
       }
    }
     
    // Add remaining possible
    // substrings consisting
    // of vowels occupying
    // last indices of the string
    if (cnt != 0)
        ans += (cnt * (cnt + 1) / 2);
 
    return ans;
}
 
// Driver code
static public void Main(string[] args)
{
    string s = "geeksforgeeks";
 
    Console.Write(CountTotal(s));
}
}
 
// This code is contributed by rock_cool

Javascript

 
    // Javascript program to count all substrings
    // in a string which contains only vowels
     
    // Function to check vowel or not
    function isvowel(x)
    {
        return (x == 'a' || x == 'e' ||
                x == 'i' || x == 'o' ||
                x == 'u');
    }
 
    // Function to find largest
    // string which satisy condition
    function CountTotal(str)
    {
        let ans = 0;
        let n = str.length;
        let s = str.split('');
        let cnt = 0;
 
        for(let i = 0; i < n; i++)
        {
 
           // Check if current character
           // is vowel
           if (isvowel(s[i]))
 
               // Increment count
               cnt += 1;
           else
           {
 
               // Count all possible
               // substrings of calculated
               // length
               ans += (cnt * (cnt + 1) / 2);
 
               // Reset the length
               cnt = 0;
           }
        }
 
        // Add remaining possible
        // substrings consisting
        // of vowels occupying
        // last indices of the string
        if (cnt != 0)
            ans += (cnt * (cnt + 1) / 2);
 
        return ans;
    }
     
    let s = "geeksforgeeks";
  
    document.write(CountTotal(s));
     

输出：

7

时间复杂度： O(N)

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