给定一个包含小写字母的字符串str ，任务是至少计算一次包含所有元音的子字符串，并且子字符串中不存在辅音(非元音字符)。
例子：
Input: str = “aeoibsddaaeiouudb”
Output: 4
“aaeiouu”, “aeiouu”, “aeiou” and “aaeiou”
Input: str = “aeoisbddiouuaedf”
Output: 1
Input: str = “aeouisddaaeeiouua”
Output: 9
方法：想法是提取仅包含元音的所有最大长度子字符串。现在，对于所有这些子字符串，我们需要至少找到一次包含所有元音的子字符串的计数。这可以使用两指针技术来完成。
在这种情况下如何使用两指针技术的说明：
If string = “aeoibsddaaeiouudb”
The first step is to extract all maximum length sub-strings that contain only vowels which are:
- aeoi
- aaeiouu
Now, take the first string “aeoi”, it will not be counted because vowel ‘u’ is missing.
Then, take the second substring i.e. “aaeiouu”
Length of the string, n = 7
start = 0
index = 0
count = 0
We will run a loop till all the vowels are present at least once, so we stop at index 5 and start = 0.
Now our string is “aaeiou” and there are n – i substrings that contain vowels at least once and have string “aaeiou” as their prefix.
These substrings are: “aaeiou”, “aaeiouu”
count = count + (n – i) = 7 – 5 = 2
Now, count = 2
Then, increment start with 1. If substring between [start, index] i.e (1, 5) still contains vowels at least once then add (n – i).
These substrings are: “aeiou”, “aeiouu”
count = count + (n – i) = 7 – 5 = 2
Now, count = 2
Then start = 2, now substring becomes “eiouu”. Then no further count can be added because vowel ‘a’ is missing.
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true if c is a vowel
bool isVowel(char c)
{
return (c == 'a' || c == 'e' || c == 'i'
|| c == 'o' || c == 'u');
}
// Function to return the count of sub-strings
// that contain every vowel at least
// once and no consonant
int countSubstringsUtil(string s)
{
int count = 0;
// Map is used to store count of each vowel
map mp;
int n = s.length();
// Start index is set to 0 initially
int start = 0;
for (int i = 0; i < n; i++) {
mp[s[i]]++;
// If substring till now have all vowels
// atleast once increment start index until
// there are all vowels present between
// (start, i) and add n - i each time
while (mp['a'] > 0 && mp['e'] > 0
&& mp['i'] > 0 && mp['o'] > 0
&& mp['u'] > 0) {
count += n - i;
mp[s[start]]--;
start++;
}
}
return count;
}
// Function to extract all maximum length
// sub-strings in s that contain only vowels
// and then calls the countSubstringsUtil() to find
// the count of valid sub-strings in that string
int countSubstrings(string s)
{
int count = 0;
string temp = "";
for (int i = 0; i < s.length(); i++) {
// If current character is a vowel then
// append it to the temp string
if (isVowel(s[i])) {
temp += s[i];
}
// The sub-string containing all vowels ends here
else {
// If there was a valid sub-string
if (temp.length() > 0)
count += countSubstringsUtil(temp);
// Reset temp string
temp = "";
}
}
// For the last valid sub-string
if (temp.length() > 0)
count += countSubstringsUtil(temp);
return count;
}
// Driver code
int main()
{
string s = "aeouisddaaeeiouua";
cout << countSubstrings(s) << endl;
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that returns true if c is a vowel
static boolean isVowel(char c)
{
return (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' || c == 'u');
}
// Function to return the count of sub-strings
// that contain every vowel at least
// once and no consonant
static int countSubstringsUtil(char []s)
{
int count = 0;
// Map is used to store count of each vowel
Map mp = new HashMap<>();
int n = s.length;
// Start index is set to 0 initially
int start = 0;
for (int i = 0; i < n; i++)
{
if(mp.containsKey(s[i]))
{
mp.put(s[i], mp.get(s[i]) + 1);
}
else
{
mp.put(s[i], 1);
}
// If substring till now have all vowels
// atleast once increment start index until
// there are all vowels present between
// (start, i) and add n - i each time
while (mp.containsKey('a') && mp.containsKey('e') &&
mp.containsKey('i') && mp.containsKey('o') &&
mp.containsKey('u') && mp.get('a') > 0 &&
mp.get('e') > 0 && mp.get('i') > 0 &&
mp.get('o') > 0 && mp.get('u') > 0)
{
count += n - i;
mp.put(s[start], mp.get(s[start]) - 1);
start++;
}
}
return count;
}
// Function to extract all maximum length
// sub-strings in s that contain only vowels
// and then calls the countSubstringsUtil() to find
// the count of valid sub-strings in that string
static int countSubstrings(String s)
{
int count = 0;
String temp = "";
for (int i = 0; i < s.length(); i++)
{
// If current character is a vowel then
// append it to the temp string
if (isVowel(s.charAt(i)))
{
temp += s.charAt(i);
}
// The sub-string containing all vowels ends here
else
{
// If there was a valid sub-string
if (temp.length() > 0)
count += countSubstringsUtil(temp.toCharArray());
// Reset temp string
temp = "";
}
}
// For the last valid sub-string
if (temp.length() > 0)
count += countSubstringsUtil(temp.toCharArray());
return count;
}
// Driver code
public static void main(String[] args)
{
String s = "aeouisddaaeeiouua";
System.out.println(countSubstrings(s));
}
}
// This code is contributed by Princi Singh Python3
# Python3 implementation of the approach
# Function that returns true if c is a vowel
def isVowel(c) :
return (c == 'a' or c == 'e' or c == 'i'
or c == 'o' or c == 'u');
# Function to return the count of sub-strings
# that contain every vowel at least
# once and no consonant
def countSubstringsUtil(s) :
count = 0;
# Map is used to store count of each vowel
mp = dict.fromkeys(s,0);
n = len(s);
# Start index is set to 0 initially
start = 0;
for i in range(n) :
mp[s[i]] += 1;
# If substring till now have all vowels
# atleast once increment start index until
# there are all vowels present between
# (start, i) and add n - i each time
while (mp['a'] > 0 and mp['e'] > 0
and mp['i'] > 0 and mp['o'] > 0
and mp['u'] > 0) :
count += n - i;
mp[s[start]] -= 1;
start += 1;
return count;
# Function to extract all maximum length
# sub-strings in s that contain only vowels
# and then calls the countSubstringsUtil() to find
# the count of valid sub-strings in that string
def countSubstrings(s) :
count = 0;
temp = "";
for i in range(len(s)) :
# If current character is a vowel then
# append it to the temp string
if (isVowel(s[i])) :
temp += s[i];
# The sub-string containing all vowels ends here
else :
# If there was a valid sub-string
if (len(temp) > 0) :
count += countSubstringsUtil(temp);
# Reset temp string
temp = "";
# For the last valid sub-string
if (len(temp) > 0) :
count += countSubstringsUtil(temp);
return count;
# Driver code
if __name__ == "__main__" :
s = "aeouisddaaeeiouua";
print(countSubstrings(s));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function that returns true if c is a vowel
static bool isVowel(char c)
{
return (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' || c == 'u');
}
// Function to return the count of sub-strings
// that contain every vowel at least
// once and no consonant
static int countSubstringsUtil(char []s)
{
int count = 0;
// Map is used to store count of each vowel
Dictionary mp = new Dictionary();
int n = s.Length;
// Start index is set to 0 initially
int start = 0;
for (int i = 0; i < n; i++)
{
if(mp.ContainsKey(s[i]))
{
mp[s[i]] = mp[s[i]] + 1;
}
else
{
mp.Add(s[i], 1);
}
// If substring till now have all vowels
// atleast once increment start index until
// there are all vowels present between
// (start, i) and add n - i each time
while (mp.ContainsKey('a') && mp.ContainsKey('e') &&
mp.ContainsKey('i') && mp.ContainsKey('o') &&
mp.ContainsKey('u') && mp['a'] > 0 &&
mp['e'] > 0 && mp['i'] > 0 &&
mp['o'] > 0 && mp['u'] > 0)
{
count += n - i;
if(mp.ContainsKey(s[start]))
mp[s[start]] = mp[s[start]] - 1;
start++;
}
}
return count;
}
// Function to extract all maximum length
// sub-strings in s that contain only vowels
// and then calls the countSubstringsUtil() to find
// the count of valid sub-strings in that string
static int countSubstrings(String s)
{
int count = 0;
String temp = "";
for (int i = 0; i < s.Length; i++)
{
// If current character is a vowel then
// append it to the temp string
if (isVowel(s[i]))
{
temp += s[i];
}
// The sub-string containing
// all vowels ends here
else
{
// If there was a valid sub-string
if (temp.Length > 0)
count += countSubstringsUtil(temp.ToCharArray());
// Reset temp string
temp = "";
}
}
// For the last valid sub-string
if (temp.Length > 0)
count += countSubstringsUtil(temp.ToCharArray());
return count;
}
// Driver code
public static void Main(String[] args)
{
String s = "aeouisddaaeeiouua";
Console.WriteLine(countSubstrings(s));
}
}
// This code is contributed by Princi Singh 输出：
9
时间复杂度： O(N)