恰好包含 K 个元音的子串计数

给定字符串str包含大写和小写字母，以及一个整数K 。任务是找出恰好包含 K 个元音（可能是重复的）的子串的数量。

例子：

Input: str = “aeiou”, K = 2
Output: 4
Explanation: The substrings are “ae”, “ei”, “io”, “ou”.

Input: str = “TrueGeek”, K = 3
Output: 5
Explanation: All such possible substrings are:
“TrueGe”, “rueGe”, “ueGe”, “eGee”, “eGeek”.

方法：解决这个问题是基于贪婪的方法。生成所有子字符串并为每个子字符串检查元音的数量。请按照以下步骤操作。

• 生成所有子字符串。对于每个子字符串，请执行以下操作
  • 存储元音的出现次数。
  • 检查子字符串中的新字符是否为元音。
  • 如果是元音，则增加找到的元音计数
  • 现在对于每个子字符串，如果元音的计数是 K ，则增加最终计数。
• 最后返回最终计数作为所需答案。

下面是上述代码的实现。

C++

// C++ code to implement above approach
#include 
using namespace std;
  
#define MAX 128
  
// Function to check whether
// a character is vowel or not
bool isVowel(char x)
{
    return (x == 'a' || x == 'e' || 
            x == 'i' || x == 'o' || x == 'u' 
            || x == 'A' || x == 'E' || 
            x == 'I' || x == 'O' || x == 'U');
}
  
// Function to find the count of
// substring with k vowels
int get(string str, int k)
{
  
    int n = str.length();
  
    // Stores the count of
    // substring with K vowels
    int ans = 0;
  
    // Consider all substrings 
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
        int count = 0;
  
        // Consider all substrings 
        // between [i, j]
        for (int j = i; j < n; j++) {
  
            // If this is a vowel, for this
            // substring, increment count.
            if (isVowel(str[j])) {
                count++;
            }
  
            // If vowel count becomes k,
            // then increment final count.
            if (count == k) {
                ans++;
            }
  
            if (count > k)
                break;
        }
    }
    return ans;
}
  
// Driver code
int main(void)
{
    string s = "aeiou";
    int K = 2;
    cout << get(s, K);
    return 0;
}

Java

// Java code to implement above approach
class GFG
{
  
  static final int MAX = 128;
  
  // Function to check whether
  // a character is vowel or not
  static boolean isVowel(char x)
  {
    return (x == 'a' || x == 'e' || 
            x == 'i' || x == 'o' || x == 'u' 
            || x == 'A' || x == 'E' || 
            x == 'I' || x == 'O' || x == 'U');
  }
  
  // Function to find the count of
  // subString with k vowels
  static int get(String str, int k)
  {
  
    int n = str.length();
  
    // Stores the count of
    // subString with K vowels
    int ans = 0;
  
    // Consider all subStrings 
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int count = 0;
  
      // Consider all subStrings 
      // between [i, j]
      for (int j = i; j < n; j++) {
  
        // If this is a vowel, for this
        // subString, increment count.
        if (isVowel(str.charAt(j))) {
          count++;
        }
  
        // If vowel count becomes k,
        // then increment final count.
        if (count == k) {
          ans++;
        }
  
        if (count > k)
          break;
      }
    }
    return ans;
  }
  
  // Driver code
  public static void main(String[] args)
  {
    String s = "aeiou";
    int K = 2;
    System.out.print(get(s, K));
  }
}
  
// This code is contributed by 29AjayKumar

Python3

# python code to implement above approach
MAX = 128
  
# Function to check whether
# a character is vowel or not
def isVowel(x):
  
    return (x == 'a' or x == 'e' or
            x == 'i' or x == 'o' or x == 'u'
            or x == 'A' or x == 'E' or
            x == 'I' or x == 'O' or x == 'U')
  
# Function to find the count of
# substring with k vowels
def get(str, k):
  
    n = len(str)
  
    # Stores the count of
    # substring with K vowels
    ans = 0
  
    # Consider all substrings
    # beginning with str[i]
    for i in range(0, n):
        count = 0
  
        # Consider all substrings
        # between [i, j]
        for j in range(i, n):
  
            # If this is a vowel, for this
            # substring, increment count.
            if (isVowel(str[j])):
                count += 1
  
            # If vowel count becomes k,
            # then increment final count.
            if (count == k):
                ans += 1
  
            if (count > k):
                break
  
    return ans
  
# Driver code
if __name__ == "__main__":
  
    s = "aeiou"
    K = 2
    print(get(s, K))
  
# This code is contributed by rakeshsahni

C#

// C# code to implement above approach
using System;
class GFG {
  
  // Function to check whether
  // a character is vowel or not
  static bool isVowel(char x)
  {
    return (x == 'a' || x == 'e' || x == 'i' || x == 'o'
            || x == 'u' || x == 'A' || x == 'E'
            || x == 'I' || x == 'O' || x == 'U');
  }
  
  // Function to find the count of
  // substring with k vowels
  static int get(string str, int k)
  {
  
    int n = str.Length;
  
    // Stores the count of
    // substring with K vowels
    int ans = 0;
  
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int count = 0;
  
      // Consider all substrings
      // between [i, j]
      for (int j = i; j < n; j++) {
  
        // If this is a vowel, for this
        // substring, increment count.
        if (isVowel(str[j])) {
          count++;
        }
  
        // If vowel count becomes k,
        // then increment final count.
        if (count == k) {
          ans++;
        }
  
        if (count > k)
          break;
      }
    }
    return ans;
  }
  
  // Driver code
  public static void Main()
  {
    string s = "aeiou";
    int K = 2;
    Console.WriteLine(get(s, K));
  }
}
  
// This code is contributed by ukasp.

Javascript

   // JavaScript code for the above approach
 
   let MAX = 128
 
   // Function to check whether
   // a character is vowel or not
   function isVowel(x) {
     return (x == 'a' || x == 'e' ||
       x == 'i' || x == 'o' || x == 'u'
       || x == 'A' || x == 'E' ||
       x == 'I' || x == 'O' || x == 'U');
   }
 
   // Function to find the count of
   // substring with k vowels
   function get(str, k) {
 
     let n = str.length;
 
     // Stores the count of
     // substring with K vowels
     let ans = 0;
 
     // Consider all substrings 
     // beginning with str[i]
     for (let i = 0; i < n; i++) {
       let count = 0;
 
       // Consider all substrings 
       // between [i, j]
       for (let j = i; j < n; j++) {
 
         // If this is a vowel, for this
         // substring, increment count.
         if (isVowel(str[j])) {
           count++;
         }
 
         // If vowel count becomes k,
         // then increment final count.
         if (count == k) {
           ans++;
         }
 
         if (count > k)
           break;
       }
     }
     return ans;
   }
 
   // Driver code
   let s = "aeiou";
   let K = 2;
   document.write(get(s, K));
 
 // This code is contributed by Potta Lokesh
 

输出

4

时间复杂度： O(N ² ) 其中 N 是字符串的长度。
辅助空间： O(1)