Count substrings that contain all vowels | SET 2
Given a string str containing lowercase alphabets, the task is to count the sub-strings that contain all the vowels at-least one time and there are no consonants (non-vowel characters) present in the sub-strings.
Examples:
Input: str = “aeoibsddaaeiouudb”
Output: 4
“aaeiouu”, “aeiouu”, “aeiou” and “aaeiou”
Input: str = “aeoisbddiouuaedf”
Output: 1
Input: str = “aeouisddaaeeiouua”
Output: 9
Approach: The idea is to extract all the maximum length sub-strings that contain only vowels. Now for all these sub-strings separately, we need to find the count of sub-strings that contains all the vowels at least once. This can be done using two-pointer technique.
Illustration of how to use the two-pointer technique in this case:
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.
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 { // 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 |
Javascript
<script> // JavaScript implementation of the approach // Function that returns true if c is a vowel function isVowel( 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 function countSubstringsUtil( s) { var count = 0; // Map is used to store count of each vowel var mp = {}; var n = s.length; // Start index is set to 0 initially var start = 0; for (let i = 0; i < n; i++) { if (mp[s[i]]){ mp[s[i]]++; } else 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 && 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 function countSubstrings( s) { var count = 0; var temp = "" ; for (let 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 var s = "aeouisddaaeeiouua" ; console.log( countSubstrings(s)); // This code is contributed by ukasp. </script> |
9
Time Complexity: O(N^2)
Auxiliary Space: O(N), since Hashmap is used and memory is allocated in each iteration.
Optimized approach using Sliding Window -:
The idea is to keep track of all the vowels and count the substrings containing all vowels using a sliding window approach.
Intution :
Maintains a count of vowels in the current substring and a temporary array to keep track of the count of each vowel in the substring.
When the count of all the vowels in the substring is 5, then count all the substring in the substring forming from i to j.
If the current character is not a vowel, it resets the start pointer and the count and the array.
Finally, it returns the answer which is the count of all substrings that contain all the vowels.
Algorithm :
1. Initialize variables i, j, count, and ans to 0, and create an integer array arr of size 26.
2. Iterate through the string using variable j.
3. If the current character ch is a vowel, increment the count of that vowel in the arr array, and if the count of that vowel is 1 (i.e., it is the first occurrence), increment the count variable.
4. If count is equal to 5 (i.e., all vowels have been seen at least once), then initialize a new integer array temp to keep track of the counts of vowels in the current substring, and loop through the characters from index i to j.
5. For each character, decrement the count of that vowel in the temp array, and if the count becomes 0, decrement the count variable.
6. Increment the ans variable for each iteration of the loop in step 5, because each substring from i to j containing all vowels will be counted.
7. Once count becomes less than 5 (i.e., the current substring no longer contains all vowels), set i to j+1, set count to 0, and create a new arr array.
8. Continue iterating through the string until the end is reached, and then return the ans variable.
Pseudocode :
function countVowelSubstrings(word):
n = length of word
arr = array of 26 zeros
i = 0
j = 0
count = 0
ans = 0
while j < n do:
ch = character at index j in word
if checkVowel(ch) then:
if arr[ch - 'a'] == 0 then:
count = count + 1
arr[ch - 'a'] = arr[ch - 'a'] + 1
if count == 5 then:
val = i
temp = array of 26 zeros
temp['a' - 'a'] = arr['a' - 'a']
temp['e' - 'a'] = arr['e' - 'a']
temp['i' - 'a'] = arr['i' - 'a']
temp['o' - 'a'] = arr['o' - 'a']
temp['u' - 'a'] = arr['u' - 'a']
while count == 5 do:
ans = ans + 1
c = character at index val in word
temp = temp - 1
if temp == 0 then:
count = count - 1
val = val + 1
count = 5
end if
else:
i = j + 1
count = 0
arr = array of 26 zeros
end if
j = j + 1
end while
return ans
end function
function checkVowel(ch):
if ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u' then:
return true
else:
return false
end if
end function
C++
#include <bits/stdc++.h> using namespace std; // function to check that a character is a vowel or not bool checkVowel( char ch) { if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' ) { return true ; } return false ; } // function will return the number of count of vowel substrings int countVowelSubstrings(string word) { int n = word.length(); int arr[26]; // initializing the array with 0 memset (arr, 0, sizeof (arr)); int i = 0; int j = 0; int count = 0; int ans = 0; while (j < n) { char ch = word[j]; if (checkVowel(ch)) { if (arr[ch - 'a' ] == 0) { count++; } arr[ch - 'a' ]++; if (count == 5) { int val = i; int temp[26]; memcpy (temp, arr, sizeof (temp)); while (count == 5) { ans++; char c = word[val]; temp--; if (temp == 0) { count--; } val++; } count = 5; } } else { i = j + 1; count = 0; memset (arr, 0, sizeof (arr)); } j++; } return ans; } // driver program to test above functions int main() { string s = "aeouisddaaeeiouua" ; cout << countVowelSubstrings(s); } |
Java
/*package whatever //do not write package name here */ import java.io.*; class GFG { public static int countVowelSubstrings(String word) { int n = word.length(); int []arr = new int [ 26 ]; int i = 0 ; int j = 0 ; int count = 0 ; int ans = 0 ; while (j < n){ char ch = word.charAt(j); if (checkVowel(ch)){ if (arr[ch - 'a' ] == 0 ){ count++; } arr[ch - 'a' ]++; if (count == 5 ){ int val = i; int []temp = new int [ 26 ]; temp[ 'a' - 'a' ] = arr[ 'a' - 'a' ]; temp[ 'e' - 'a' ] = arr[ 'e' - 'a' ]; temp[ 'i' - 'a' ] = arr[ 'i' - 'a' ]; temp[ 'o' - 'a' ] = arr[ 'o' - 'a' ]; temp[ 'u' - 'a' ] = arr[ 'u' - 'a' ]; while (count == 5 ){ ans++; char c = word.charAt(val); temp--; if (temp == 0 ){ count--; } val++; } count = 5 ; } } else { i = j+ 1 ; count = 0 ; arr = new int [ 26 ]; } j++; } return ans; } public static boolean checkVowel( char ch){ if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' ){ return true ; } return false ; } public static void main (String[] args) { String s = "aeouisddaaeeiouua" ; System.out.println(countVowelSubstrings(s)); } } // This code is contributed by vishalkumarsahu04 |
Python
# Function to check if a character is a vowel or not def check_vowel(ch): return ch in [ 'a' , 'e' , 'i' , 'o' , 'u' ] # Function to count vowel substrings def count_vowel_substrings(word): n = len (word) arr = [ 0 ] * 26 # Initializing the array with 0 i = j = count = ans = 0 while j < n: ch = word[j] if check_vowel(ch): if arr[ ord (ch) - ord ( 'a' )] = = 0 : count + = 1 arr[ ord (ch) - ord ( 'a' )] + = 1 if count = = 5 : val = i temp = arr[:] while count = = 5 : ans + = 1 c = word[val] temp[ ord (c) - ord ( 'a' )] - = 1 if temp[ ord (c) - ord ( 'a' )] = = 0 : count - = 1 val + = 1 count = 5 else : i = j + 1 count = 0 arr = [ 0 ] * 26 j + = 1 return ans # Driver program to test the above functions if __name__ = = "__main__" : s = "aeouisddaaeeiouua" print (count_vowel_substrings(s)) |
C#
using System; class Program { // function to check that a character is a vowel or not static bool CheckVowel( char ch) { if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' ) { return true ; } return false ; } // function will return the number of count of vowel // substrings static int CountVowelSubstrings( string word) { int n = word.Length; int [] arr = new int [26]; // initializing the array with 0 Array.Fill(arr, 0); int i = 0; int j = 0; int count = 0; int ans = 0; while (j < n) { char ch = word[j]; if (CheckVowel(ch)) { if (arr[ch - 'a' ] == 0) { count++; } arr[ch - 'a' ]++; if (count == 5) { int val = i; int [] temp = new int [26]; Array.Copy(arr, temp, arr.Length); while (count == 5) { ans++; char c = word[val]; temp--; if (temp == 0) { count--; } val++; } count = 5; } } else { i = j + 1; count = 0; Array.Fill(arr, 0); } j++; } return ans; } // driver program to test above functions static void Main( string [] args) { string s = "aeouisddaaeeiouua" ; Console.WriteLine(CountVowelSubstrings(s)); } } // This code is contributed by user_dtewbxkn77n |
Javascript
// Function to check if a character is a vowel or not function checkVowel(ch) { return ch === 'a' || ch === 'e' || ch === 'i' || ch === 'o' || ch === 'u' ; } // Function to count vowel substrings function countVowelSubstrings(word) { const n = word.length; const arr = Array(26).fill(0); let i = 0; let j = 0; let count = 0; let ans = 0; while (j < n) { const ch = word[j]; if (checkVowel(ch)) { if (arr[ch.charCodeAt(0) - 'a' .charCodeAt(0)] === 0) { count++; } arr[ch.charCodeAt(0) - 'a' .charCodeAt(0)]++; if (count === 5) { let val = i; const temp = [...arr]; while (count === 5) { ans++; const c = word[val]; temp--; if (temp === 0) { count--; } val++; } count = 5; } } else { i = j + 1; count = 0; arr.fill(0); } j++; } return ans; } // Driver program to test the function function main() { const s = "aeouisddaaeeiouua" ; console.log(countVowelSubstrings(s)); } // Run the main function main(); |
9
Time Complexity — O(N)
Auxiliary Space — O(26) i.e O(1)