计算给定字符串的所有子字符串中出现的元音数

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，任务是找到在给定字符串 的所有子字符串中出现的元音数。
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

简单方法：给定的长度N，可以形成子串的数目= N（N + 1）/ 2的字符串。一个简单的解决方案是对于每个子字符串，我们计算元音的出现次数并将它们相加以获得结果。这种方法的时间复杂度为 O(N ³ )，不适用于较大的 N 值。
有效的方法：这个想法是使用基于前缀和数组的技术，我们将每个字符在所有连接的子字符串中的出现次数存储起来。

• 对于第一个字符，

no. of occurrences = no. of substrings starting with the first character = N.

• 对于以下每个字符，我们存储

no. of substrings starting with that character + the number of substrings formed by the previous characters containing this character – the number of substrings formed by the previous characters only.

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Returns the total sum of
// occurrences of all vowels
int vowel_calc(string s)
{
    int n = s.length();
    vector arr;
 
    for (int i = 0; i < n; i++) {
 
        if (i == 0)
            // No. of occurrences of 0th character
            // in all the substrings
            arr.push_back(n);
 
        else
            // No. of occurrences of the ith character
            // in all the substrings
            arr.push_back((n - i) + arr[i - 1] - i);
    }
 
    int sum = 0;
    for (int i = 0; i < n; i++)
 
        // Check if ith character is a vowel
        if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
            || s[i] == 'o' || s[i] == 'u')
            sum += arr[i];
 
    // Return the total sum
    // of occurrences of vowels
    return sum;
}
 
// Driver code
int main()
{
    string s = "daceh";
    cout << vowel_calc(s) << endl;
 
    return 0;
}

Java

// Java implementation of the above approach
 
import java.io.*;
import java.util.*;
 
public class Gfg {
 
    // Returns the total sum of
    // occurrences of all vowels
    static int vowel_calc(String s)
    {
        int n = s.length();
        int arr[] = new int[n];
 
        for (int i = 0; i < n; i++) {
 
            if (i == 0)
                // No. of occurrences of 0th character
                // in all the substrings
                arr[i] = n;
 
            else
                // No. of occurrences of ith character
                // in all the substrings
                arr[i] = (n - i) + arr[i - 1] - i;
        }
 
        int sum = 0;
        for (int i = 0; i < n; i++) {
            char ch = s.charAt(i);
            // Check if ith character is a vowel
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u')
                sum += arr[i];
        }
 
        // Return the total sum
        // of occurrences of vowels
        return sum;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String s = "daceh";
        System.out.println(vowel_calc(s));
    }
}

Python3

# Python 3 implementation of
# a more efficient approach.
# return sum of all occurrences of all vowels
def sumVowel(string):
    n = len(string)
    sum = 0
    string = string.lower()
 
    # iterate through every character in the string
    for i in range(0, n):
        s = string[i]
 
        # checks if the character is a vowel or not
        if (s=="a" or s == "e" or s == "i" or s == "o" or s == "u"):
 
            # uses below expression to calculate the count
                        # of all occurrences of character in substrings
                        # of the string
            sum += ((n - i) * (i + 1))           
 
    # return the total sum of occurrence
    return sum
 
#driver code
if __name__ == '__main__':
    #input string here
    string = "abhay"
    #print returned sum
    print(sumVowel(string))
 
# This code is contributed by
# Abhay Subramanian K

C#

// C# implementation of the above approach
  
 
using System;
  
public class Gfg {
  
    // Returns the total sum of
    // occurrences of all vowels
    static int vowel_calc(string s)
    {
        int n = s.Length;
        int[] arr = new int[n];
  
        for (int i = 0; i < n; i++) {
  
            if (i == 0)
                // No. of occurrences of 0th character
                // in all the substrings
                arr[i] = n;
  
            else
                // No. of occurrences of ith character
                // in all the substrings
                arr[i] = (n - i) + arr[i - 1] - i;
        }
  
        int sum = 0;
        for (int i = 0; i < n; i++) {
            char ch = s[i];
            // Check if ith character is a vowel
            if (ch == 'a' || ch == 'e' || ch == 'i'
                || ch == 'o' || ch == 'u')
                sum += arr[i];
        }
  
        // Return the total sum
        // of occurrences of vowels
        return sum;
    }
  
    // Driver Code
    public static void Main()
    {
        string s = "daceh";
        Console.Write(vowel_calc(s));
    }
}

Javascript

 
// Javascript implementation of the above approach
 
// Returns the total sum of
// occurrences of all vowels
function vowel_calc(s)
{
    var n = s.length;
    var arr = [];
 
    for (var i = 0; i < n; i++) {
 
        if (i == 0)
            // No. of occurrences of 0th character
            // in all the substrings
            arr.push(n);
 
        else
            // No. of occurrences of the ith character
            // in all the substrings
            arr.push((n - i) + arr[i - 1] - i);
    }
 
    var sum = 0;
    for (var i = 0; i < n; i++)
 
        // Check if ith character is a vowel
        if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i'
            || s[i] == 'o' || s[i] == 'u')
            sum += arr[i];
 
    // Return the total sum
    // of occurrences of vowels
    return sum;
}
 
// Driver code
var s = "daceh";
document.write( vowel_calc(s));
 
// This code is contributed by importantly.

输出：

16

时间复杂度： O(N)