Подсчитать подстроки, содержащие все гласные | НАБОР 2

Опубликовано: 7 Января, 2022

Учитывая строку str, содержащую строчные буквы, задача состоит в том, чтобы подсчитать подстроки, которые содержат все гласные хотя бы один раз, и в подстроках нет согласных (негласных символов).

Примеры:

Input: str = “aeoibsddaaeiouudb”
Output: 4
“aaeiouu”, “aeiouu”, “aeiou” and “aaeiou”

Input: str = “aeoisbddiouuaedf”
Output: 1

Input: str = “aeouisddaaeeiouua”
Output: 9

Рекомендуется: сначала попробуйте свой подход в {IDE}, прежде чем переходить к решению.

Подход: идея состоит в том, чтобы извлечь все подстроки максимальной длины, содержащие только гласные. Теперь для всех этих подстрок по отдельности нам нужно найти количество подстрок, которые содержат все гласные хотя бы один раз. Это можно сделать, используя технику двух указателей.

Иллюстрация того, как использовать технику двух указателей в этом случае:

If string = “aeoibsddaaeiouudb”
The first step is to extract all maximum length sub-strings that contain only vowels which are:

  1. aeoi
  2. 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.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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<char, int> 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<Character, Integer> 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 AnkitRai01

C#

// 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<char,
               int> mp = new Dictionary<char
                                        int>();
  
    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