计算给定字符串的所有子字符串中出现的元音数 |设置 2

给定一个长度为N的包含0个或多个元音的小写字符的字符串str[] ，任务是找出在给定字符串的所有子字符串中出现的元音计数。

例子：

Input: str = “abc” 
Output: 3
The given string “abc” contains only one vowel = ‘a’ 
Substrings of “abc” are = {“a”, “b”, “c”, “ab”, “bc, “abc”} 
Hence, the sum of occurrences of the vowel in these strings = 3.(‘a’ occurred 3 times)

Input: str = “daceh” 
Output: 16

前缀求和方法：本文的第 1 组讨论了朴素方法和使用前缀求和的方法。

有效方法：假设给定字符串是str ，子字符串从位置x开始，到位置y结束，如果元音在第 i 个位置，其中0<=x<=i和i<=y<-N 。对于每个元音，它可以在子字符串中，即 x 的(i+1)个选项和 y的(ni)个选项。所以总共有(i+1)*(ni) 个包含str[i] 的子串。取一个常数“ aeiou ”。请按照以下步骤解决问题：

• 将变量res初始化为0以存储答案。
• 使用变量i遍历范围[0, N)并执行以下任务：
  • 如果str[i]是元音，则将(I+1)*(Ni)的值添加到变量res以计算可能的元音总数。
• 执行上述步骤后，打印res的值作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the number of vowels
long long countVowels(string str)
{
 
    // Define the size of the string
    // and ans as zero
    long res = 0, N = str.size();
 
    // Start traversing and find if their
    // is any vowel or not
    for (int i = 0; i < N; ++i)
 
        // If there is vowel, then count
        // the numbers of vowels
        if (string("aeiou").find(str[i])
            != string::npos)
            res += (i + 1) * (N - i);
 
    return res;
}
 
// Driver Code
int main()
{
    string str = "daceh";
 
    // Call the function
    long long ans = countVowels(str);
 
    cout << ans << endl;
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
  // Function to count the number of vowels
  static long countVowels(String str)
  {
 
    // Define the size of the String
    // and ans as zero
    long res = 0, N = str.length();
 
    // Start traversing and find if their
    // is any vowel or not
    for (int i = 0; i < N; ++i)
 
      // If there is vowel, then count
      // the numbers of vowels
      if (new String("aeiou").contains(String.valueOf(str.charAt(i))))
        res += (i + 1) * (N - i);
 
    return res;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String str = "daceh";
 
    // Call the function
    long ans = countVowels(str);
    System.out.print(ans +"\n");
 
  }
}
 
// This code is contributed by shikhasingrajput.

Python3

# Python program for the above approach
 
# Function to count the number of vowels
def countVowels(str):
   
    # Define the size of the String
    # and ans as zero
    res = 0;
    N = len(str);
 
    # Start traversing and find if their
    # is any vowel or not
    for i in range(N):
 
        # If there is vowel, then count
        # the numbers of vowels
        if ((str[i]) in ("aeiou")):
            res += (i + 1) * (N - i);
    return res;
 
# Driver Code
if __name__ == '__main__':
 
    str = "daceh";
 
    # Call the function
    ans = countVowels(str);
    print(ans);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
class GFG{
 
  // Function to count the number of vowels
  static long countVowels(String str)
  {
 
    // Define the size of the String
    // and ans as zero
    long res = 0, N = str.Length;
 
    // Start traversing and find if their
    // is any vowel or not
    for (int i = 0; i < N; ++i)
 
      // If there is vowel, then count
      // the numbers of vowels
      if (new String("aeiou").Contains(str[i]))
        res += (i + 1) * (N - i);
 
    return res;
  }
 
  // Driver Code
  public static void Main()
  {
    String str = "daceh";
 
    // Call the function
    long ans = countVowels(str);
    Console.Write(ans +"\n");
 
  }
}
 
// This code is contributed by gfgking

Javascript

// javascript program for the above approach
// Function to count the number of vowels
  function countVowels(str)
  {
 
    // Define the size of the String
    // and ans as zero
    var res = 0, N = str.length;
 
    // Start traversing and find if their
    // is any vowel or not
    for (var i = 0; i < N; ++i)
 
      // If there is vowel, then count
      // the numbers of vowels
      if ("aeiou".includes(str[i]))
        res += (i + 1) * (N - i);
 
    return res;
  }
 
// Driver Code
var str = "daceh";
 
// Call the function
var ans = countVowels(str);
document.write(ans);
 
// This code is contributed by shikhasingrajput

输出

16

时间复杂度： O(N)
辅助空间： O(1)