以元音开头并以辅音结尾的子串，反之亦然

给定一个字符串s，计算其中的特殊子字符串。如果满足以下任一属性，则称 S 的子串是特殊的。

• 它以元音开头，以辅音结尾。
• 它以辅音开头，以元音结尾。

例子：

Input : S = "aba"
Output : 2
Substrings of S are : a, ab, aba, b, ba, a 
Out of these only 'ab' and 'ba' satisfy the
condition for special Substring. So the 
answer is 2.

Input : S = "adceba"
Output : 9

一个简单的解决方案是生成所有子字符串。对于每个子字符串，检查特殊字符串的条件。如果是，则增加计数。
一个有效的解决方案是计算字符串的每个后缀中的元音和辅音。数完这些，我们从头开始遍历字符串。对于每个辅音，我们在其后添加元音数来得到结果。同样，对于每个元音，我们在其后添加辅音数量。

C++

// CPP program to count special strings
#include 
using namespace std;
 
// Returns true if ch is vowel
bool isVowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// function to check consonant
bool isCons(char ch)
{
    return (ch != 'a' && ch != 'e' &&
            ch != 'i' && ch != 'o' &&
            ch != 'u');
}
 
int countSpecial(string &str)
{
    int len = str.length();
       
      //in case of empty string, we can't fulfill the
      //required condition, hence we return ans as 0.
      if(len == 0)
      return 0;
 
    // co[i] is going to store counts
    // of consonants from str[len-1]
    // to str[i].
    // vo[i] is going to store counts
    // of vowels from str[len-1]
    // to str[i].
    int co[len + 1];
    int vo[len + 1];
    memset(co, 0, sizeof(co));
    memset(vo, 0, sizeof(vo));
 
    // Counting consonants and vowels
    // from end of string.
    if (isCons(str[len - 1]) == 1)
        co[len-1] = 1;
    else
        vo[len-1] = 1;
    for (int i = len-2; i >= 0; i--)
    {
        if (isCons(str[i]) == 1)
        {
            co[i] = co[i + 1] + 1;
            vo[i] = vo[i + 1];
        }
        else
        {
            co[i] = co[i + 1];
            vo[i] = vo[i + 1] + 1;
        }
    }
 
    // Now we traverse string from beginning
    long long ans = 0;
    for (int i = 0; i < len; i++)
    {
        // If vowel, then count of substrings
        // starting with str[i] is equal to
        // count of consonants after it.
        if (isVowel(str[i]))
           ans = ans + co[i + 1];
 
        // If consonant, then count of
        // substrings starting with str[i]
        // is equal to count of vowels
        // after it.
        else
           ans = ans + vo[i + 1];
    }
 
    return ans;
}
 
// driver program
int main()
{
    string str = "adceba";
    cout << countSpecial(str);
    return 0;
}

Java

// Java program to count special strings
class GfG
{
 
// Returns true if ch is vowel
static boolean isVowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// function to check consonant
static boolean isCons(char ch)
{
    return (ch != 'a' && ch != 'e' &&
            ch != 'i' && ch != 'o' &&
            ch != 'u');
}
 
static int countSpecial(char []str)
{
    int len = str.length;
   
      //in case of empty string, we can't fullfill the
      //required condition, hence we return ans as 0.
      if(len == 0)
          return 0;
 
    // co[i] is going to store counts
    // of consonants from str[len-1]
    // to str[i].
    // vo[i] is going to store counts
    // of vowels from str[len-1]
    // to str[i].
    int co[] = new int[len + 1];
    int vo[] = new int[len + 1];
 
 
    // Counting consonants and vowels
    // from end of string.
    if (isCons(str[len - 1]) == true)
        co[len-1] = 1;
    else
        vo[len-1] = 1;
    for (int i = len-2; i >= 0; i--)
    {
        if (isCons(str[i]) == true)
        {
            co[i] = co[i + 1] + 1;
            vo[i] = vo[i + 1];
        }
        else
        {
            co[i] = co[i + 1];
            vo[i] = vo[i + 1] + 1;
        }
    }
 
    // Now we traverse string from beginning
    long ans = 0;
    for (int i = 0; i < len; i++)
    {
        // If vowel, then count of substrings
        // starting with str[i] is equal to
        // count of consonants after it.
        if (isVowel(str[i]))
        ans = ans + co[i + 1];
 
        // If consonant, then count of
        // substrings starting with str[i]
        // is equal to count of vowels
        // after it.
        else
        ans = ans + vo[i + 1];
    }
 
    return (int) ans;
}
 
// Driver program
public static void main(String[] args)
{
    String str = "adceba";
    System.out.println(countSpecial(str.toCharArray()));
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 program to count special strings
 
# Returns true if ch is vowel
def isVowel(ch):
    return (ch == 'a' or ch == 'e' or
            ch == 'i' or ch == 'o' or
            ch == 'u')
 
# Function to check consonant
def isCons( ch):
    return (ch != 'a' and ch != 'e' and
            ch != 'i' and ch != 'o' and
            ch != 'u')
 
def countSpecial(str):
    lent = len(str)
     
    #in case of empty string, we can't fullfill the
      #required condition, hence we return ans as 0.
      if lent == 0:
      return 0;
     
    # co[i] is going to store counts
    # of consonants from str[len-1]
    # to str[i].
    # vo[i] is going to store counts
    # of vowels from str[len-1]
    # to str[i].
    co = []
    vo = []
     
    for i in range(0, lent + 1):
        co.append(0)
         
    for i in range(0, lent + 1):
        vo.append(0)
     
    # Counting consonants and vowels
    # from end of string.
    if isCons(str[lent - 1]) == 1:
        co[lent-1] = 1
    else:
        vo[lent - 1] = 1
         
    for i in range(lent-2, -1,-1):
         
        if isCons(str[i]) == 1:
            co[i] = co[i + 1] + 1
            vo[i] = vo[i + 1]
             
        else:
            co[i] = co[i + 1]
            vo[i] = vo[i + 1] + 1
 
    # Now we traverse string from beginning
    ans = 0
     
    for i in range(lent):
         
        #If vowel, then count of substrings
        # starting with str[i] is equal to
        # count of consonants after it.
        if isVowel(str[i]):
            ans = ans + co[i + 1]
 
        #If consonant, then count of
        # substrings starting with str[i]
        # is equal to count of vowels
        # after it.
        else:
            ans = ans + vo[i + 1]
 
    return ans
 
# Driver Code
str = "adceba"
print(countSpecial(str))
 
# This code is contributed by Upendra singh bartwal

C#

// C# program to count special strings
using System;
 
class GFG
{
 
// Returns true if ch is vowel
static Boolean isVowel(char ch)
{
    return (ch == 'a' || ch == 'e' ||
            ch == 'i' || ch == 'o' ||
            ch == 'u');
}
 
// function to check consonant
static Boolean isCons(char ch)
{
    return (ch != 'a' && ch != 'e' &&
            ch != 'i' && ch != 'o' &&
            ch != 'u');
}
 
static int countSpecial(char []str)
{
    int len = str.Length;
   
      //in case of empty string, we can't fullfill the
      //required condition, hence we return ans as 0.
      if(len == 0)
          return 0;
 
    // co[i] is going to store counts
    // of consonants from str[len-1]
    // to str[i].
    // vo[i] is going to store counts
    // of vowels from str[len-1]
    // to str[i].
    int []co = new int[len + 1];
    int []vo = new int[len + 1];
 
    // Counting consonants and vowels
    // from end of string.
    if (isCons(str[len - 1]) == true)
        co[len - 1] = 1;
    else
        vo[len - 1] = 1;
    for (int i = len - 2; i >= 0; i--)
    {
        if (isCons(str[i]) == true)
        {
            co[i] = co[i + 1] + 1;
            vo[i] = vo[i + 1];
        }
        else
        {
            co[i] = co[i + 1];
            vo[i] = vo[i + 1] + 1;
        }
    }
 
    // Now we traverse string from beginning
    long ans = 0;
    for (int i = 0; i < len; i++)
    {
        // If vowel, then count of substrings
        // starting with str[i] is equal to
        // count of consonants after it.
        if (isVowel(str[i]))
        ans = ans + co[i + 1];
 
        // If consonant, then count of
        // substrings starting with str[i]
        // is equal to count of vowels
        // after it.
        else
        ans = ans + vo[i + 1];
    }
    return (int) ans;
}
 
// Driver program
public static void Main(String[] args)
{
    String str = "adceba";
    Console.WriteLine(countSpecial(str.ToCharArray()));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program to count special strings
// Returns true if ch is vowel
function isVowel(ch)
{
    return(ch === "a" || ch === "e" ||
           ch === "i" || ch === "o" ||
           ch === "u");
}
 
// Function to check consonant
function isCons(ch)
{
    return(ch !== "a" && ch !== "e" &&
           ch !== "i" && ch !== "o" &&
           ch !== "u");
}
 
function countSpecial(str)
{
    var len = str.length;
     
    //in case of empty string, we can't fullfill the
      //required condition, hence we return ans as 0.
    if(len == 0)
        return 0;
     
    // co[i] is going to store counts
    // of consonants from str[len-1]
    // to str[i].
    // vo[i] is going to store counts
    // of vowels from str[len-1]
    // to str[i].
    var co = new Array(len + 1).fill(0);
    var vo = new Array(len + 1).fill(0);
     
    // Counting consonants and vowels
    // from end of string.
    if (isCons(str[len - 1]) === true) co[len - 1] = 1;
    else vo[len - 1] = 1;
     
    for(var i = len - 2; i >= 0; i--)
    {
        if (isCons(str[i]) === true)
        {
            co[i] = co[i + 1] + 1;
            vo[i] = vo[i + 1];
        }
        else
        {
            co[i] = co[i + 1];
            vo[i] = vo[i + 1] + 1;
        }
    }
     
    // Now we traverse string from beginning
    var ans = 0;
    for(var i = 0; i < len; i++)
    {
         
        // If vowel, then count of substrings
        // starting with str[i] is equal to
        // count of consonants after it.
        if (isVowel(str[i])) ans = ans + co[i + 1];
         
        // If consonant, then count of
        // substrings starting with str[i]
        // is equal to count of vowels
        // after it.
        else ans = ans + vo[i + 1];
    }
    return parseInt(ans);
}
 
// Driver code
var str = "adceba";
 
document.write(countSpecial(str.split("")));
 
// This code is contributed by rdtank
 

输出

9

时间复杂度： O( |str| ) ，因为我们进行线性遍历来计算元音和辅音的数量。

上述解决方案的空间复杂度： O(|str|)，因为我们制作了两个数组 vo[] 和 co[] 来存储任何索引 i 处的元音数量。

通过仅使用两个变量而不是制作两个数组来存储任何索引 i 处的元音和辅音，可以优化上述解决方案以在O(1) 空间复杂度内完成。以下代码描述了相同的内容：

C++

#include 
using namespace std;
 
// function to check if a character is vowel or not
bool isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
        || ch == 'u')
        return true;
    return false;
}
 
long long countSpecial(string str)
{
    long long cnt = 0;
    int n = str.size();
 
    // in case of single character or empty string
    // we can't fulfil the given condition , hence the
    // count is 0.
    if (n == 1 || n == 0)
        return 0;
 
    // variables to store count of total vowels and
    // consonants in the string
    long long vow = 0, cons = 0;
 
    for (int i = 0; i < n; i++)
        vow += isVowel(str[i]);
    cons = n - vow;
 
    for (int i = 0; i < n; i++) {
 
        // as we encounter a vowel, we add no. of consonants
        // after it to our answer and decrease the value of
        // vow by 1, indicating that now the remaining
        // string has one vowel less than current string
        if (isVowel(str[i])) {
            vow--;
            cnt += cons;
        }
 
        // same case as above for consonants
        else {
            cons--;
            cnt += vow;
        }
    }
 
    // finally we return the cnt as our answer
    return cnt;
}
 
int main()
{
    string str = "adceba";
    cout << countSpecial(str);
    return 0;
}

C

#include 
 
// function to check if a character is vowel or not
int isVowel(char ch)
{
    if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
        || ch == 'u')
        return 1;
    return 0;
}
 
long long countSpecial(char str[], int n)
{
    long long cnt = 0;
 
    // in case of single character or empty string
    // we can't fulfil the given condition , hence the
    // count is 0.
    if (n == 1 || n == 0)
        return 0;
 
    // variables to store count of total vowels and
    // consonants in the string
    long long vow = 0, cons = 0;
 
    for (int i = 0; i < n; i++)
        vow += isVowel(str[i]);
    cons = n - vow;
 
    for (int i = 0; i < n; i++) {
        // as we encounter a vowel, we add no. of consonants
        // after it to our answer and decrease the value of
        // vow by 1, indicating that now the remaining
        // string has one vowel less than current string
        if (isVowel(str[i])) {
            vow--;
            cnt += cons;
        }
        // same case as above for consonants
        else {
            cons--;
            cnt += vow;
        }
    }
 
    // finally we return the cnt as our answer
    return cnt;
}
 
int main()
{
 
    char str[] = { 'a', 'd', 'c', 'e', 'b', 'a' };
    int n = sizeof(str) / sizeof(char);
 
    if (n == 0) {
        printf("0");
    }
    else {
 
        long long count = countSpecial(str, n);
 
        printf("%lld", count);
    }
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    // function to check if a character is vowel or not
    public static int isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
            || ch == 'u')
            return 1;
        return 0;
    }
 
    public static long countSpecial(String str)
    {
        long cnt = 0;
        int n = str.length();
 
        // in case of single character or empty string
        // we can't fulfil the given condition , hence the
        // count is 0.
        if (n == 1 || n == 0)
            return 0;
 
        // variables to store count of total vowels and
        // consonants in the string
        long vow = 0, cons = 0;
 
        for (int i = 0; i < n; i++) {
            char ch = str.charAt(i);
            vow += isVowel(ch);
        }
        cons = n - vow;
 
        for (int i = 0; i < n; i++) {
            char ch = str.charAt(i);
            // as we encounter a vowel, we add no. of
            // consonants after it to our answer
            // and decrease the value of vow by 1,
            // indicating that now the remaining
            // string has one vowel less than current string
            if (isVowel(ch) == 1) {
                vow--;
                cnt += cons;
            }
            // same case as above for consonants
            else {
                cons--;
                cnt += vow;
            }
        }
 
        // finally we return the cnt as our answer
        return cnt;
    }
 
    public static void main(String[] args)
    {
        String str = "adceba";
        long count = countSpecial(str);
        System.out.println(count);
    }
}

Python3

'''package whatever #do not write package name here '''
 
# function to check if a character is vowel or not
def isVowel(ch):
    if (ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u'):
        return 1;
    return 0;
 
def countSpecial( str):
    cnt = 0;
    n = len(str);
 
    # in case of single character or empty string
    # we can't fulfil the given condition , hence the
    # count is 0.
    if (n == 1 or n == 0):
        return 0;
 
    # variables to store count of total vowels and
    # consonants in the string
    vow = 0
    cons = 0;
 
    for i in range(n):
        ch = str[i];
        vow += isVowel(ch);
     
    cons = n - vow;
 
    for i in range(n):
        ch = str[i];
        # as we encounter a vowel, we add no. of
        # consonants after it to our answer
        # and decrease the value of vow by 1,
        # indicating that now the remaining
        # string has one vowel less than current string
        if (isVowel(ch) == 1):
            vow -= 1;
            cnt += cons;
         
        # same case as above for consonants
        else:
            cons -= 1;
            cnt += vow;
         
    # finally we return the cnt as our answer
    return cnt;
 
# Driver code
if __name__ == '__main__':
    str = "adceba";
    count = countSpecial(str);
    print(count);
 
# This code is contributed by Rajput-Ji

C＃

/*package whatever //do not write package name here */
using System;
class GFG
{
   
    // function to check if a character is vowel or not
    public static int isVowel(char ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
            || ch == 'u')
            return 1;
        return 0;
    }
 
    public static long countSpecial(String str)
    {
        long cnt = 0;
        int n = str.Length;
 
        // in case of single character or empty string
        // we can't fulfil the given condition , hence the
        // count is 0.
        if (n == 1 || n == 0)
            return 0;
 
        // variables to store count of total vowels and
        // consonants in the string
        long vow = 0, cons = 0;
 
        for (int i = 0; i < n; i++) {
            char ch = str[i];
            vow += isVowel(ch);
        }
        cons = n - vow;
 
        for (int i = 0; i < n; i++) {
            char ch = str[i];
           
            // as we encounter a vowel, we add no. of
            // consonants after it to our answer
            // and decrease the value of vow by 1,
            // indicating that now the remaining
            // string has one vowel less than current string
            if (isVowel(ch) == 1) {
                vow--;
                cnt += cons;
            }
           
            // same case as above for consonants
            else {
                cons--;
                cnt += vow;
            }
        }
 
        // finally we return the cnt as our answer
        return cnt;
    }
 
  // Driver code
    public static void Main(String[] args)
    {
        string str = "adceba";
        long count = countSpecial(str);
        Console.Write(count);
    }
}
 
// This code is contributed by shivanisinghss2110

Javascript

/*package whatever //do not write package name here */
 
// function to check if a character is vowel or not
function isVowel(ch)
    {
        if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o'
            || ch == 'u')
            return 1;
        return 0;
    }
 
    function countSpecial(str)
    {
        var cnt = 0;
        var n = str.length;
 
        // in case of single character or empty string
        // we can't fulfil the given condition , hence the
        // count is 0.
        if (n == 1 || n == 0)
            return 0;
 
        // variables to store count of total vowels and
        // consonants in the string
        var vow = 0, cons = 0;
 
        for (var i = 0; i < n; i++) {
            var ch = str.charAt(i);
            vow += isVowel(ch);
        }
        cons = n - vow;
 
        for (var i = 0; i < n; i++) {
            var ch = str.charAt(i);
            // as we encounter a vowel, we add no. of
            // consonants after it to our answer
            // and decrease the value of vow by 1,
            // indicating that now the remaining
            // string has one vowel less than current string
            if (isVowel(ch) == 1) {
                vow--;
                cnt += cons;
            }
            // same case as above for consonants
            else {
                cons--;
                cnt += vow;
            }
        }
 
        // finally we return the cnt as our answer
        return cnt;
    }
 
        var str = "adceba";
        var count = countSpecial(str);
        document.write(count);
  
// This code is contributed by shivanisinghss2110

输出

9

时间复杂度： O( |str| )

空间复杂度： O(1)