Given a string str[] of length N of lowercase characters containing 0 or more vowels, the task is to find the count of vowels that occurred in all the substrings of the given string.
Examples:
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
Prefix Sum Approach: The Naive Approach and the approach using prefix sum are discussed in the Set 1 of this article.
Efficient Approach: Suppose, given string is str, substring starts from position x, and ends at position y and if the vowel are at i-th position, where 0<=x<=i and i<=y<-N . for each vowel, it could be in the substring, that is (i+1) choices for x and (n-i) choices for y. So there are total (i+1)*(n-i) substrings containing str[i]. Take a constant “aeiou“. Follow the steps below to solve the problem:
- Initialize the variable res as 0 to store the answer.
-
Iterate over the range [0, N) using the variable i and perform the following tasks:
- If str[i] is a vowel then add the value of (I+1)*(N-i) to the variable res to count the total number of vowels possible.
- After performing the above steps, print the value of res as the answer.
Below is the implementation of the above approach.
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count the number of vowels long long countVowels(string str)
{ // Define the size of the string
// and ans as zero
long res = 0, N = str.size();
// Start traversing and find if their
// is any vowel or not
for ( int i = 0; i < N; ++i)
// If there is vowel, then count
// the numbers of vowels
if (string( "aeiou" ).find(str[i])
!= string::npos)
res += (i + 1) * (N - i);
return res;
} // Driver Code int main()
{ string str = "daceh" ;
// Call the function
long long ans = countVowels(str);
cout << ans << endl;
return 0;
} |
// Java program for the above approach class GFG{
// Function to count the number of vowels
static long countVowels(String str)
{
// Define the size of the String
// and ans as zero
long res = 0 , N = str.length();
// Start traversing and find if their
// is any vowel or not
for ( int i = 0 ; i < N; ++i)
// If there is vowel, then count
// the numbers of vowels
if ( new String( "aeiou" ).contains(String.valueOf(str.charAt(i))))
res += (i + 1 ) * (N - i);
return res;
}
// Driver Code
public static void main(String[] args)
{
String str = "daceh" ;
// Call the function
long ans = countVowels(str);
System.out.print(ans + "\n" );
}
} // This code is contributed by shikhasingrajput. |
# Python program for the above approach # Function to count the number of vowels def countVowels( str ):
# Define the size of the String
# and ans as zero
res = 0 ;
N = len ( str );
# Start traversing and find if their
# is any vowel or not
for i in range (N):
# If there is vowel, then count
# the numbers of vowels
if (( str [i]) in ( "aeiou" )):
res + = (i + 1 ) * (N - i);
return res;
# Driver Code if __name__ = = '__main__' :
str = "daceh" ;
# Call the function
ans = countVowels( str );
print (ans);
# This code is contributed by 29AjayKumar |
// C# program for the above approach using System;
class GFG{
// Function to count the number of vowels
static long countVowels(String str)
{
// Define the size of the String
// and ans as zero
long res = 0, N = str.Length;
// Start traversing and find if their
// is any vowel or not
for ( int i = 0; i < N; ++i)
// If there is vowel, then count
// the numbers of vowels
if ( new String( "aeiou" ).Contains(str[i]))
res += (i + 1) * (N - i);
return res;
}
// Driver Code
public static void Main()
{
String str = "daceh" ;
// Call the function
long ans = countVowels(str);
Console.Write(ans + "\n" );
}
} // This code is contributed by gfgking |
<script> // javascript program for the above approach // Function to count the number of vowels function countVowels(str)
{
// Define the size of the String
// and ans as zero
var res = 0, N = str.length;
// Start traversing and find if their
// is any vowel or not
for ( var i = 0; i < N; ++i)
// If there is vowel, then count
// the numbers of vowels
if ( "aeiou" .includes(str[i]))
res += (i + 1) * (N - i);
return res;
}
// Driver Code var str = "daceh" ;
// Call the function var ans = countVowels(str);
document.write(ans); // This code is contributed by shikhasingrajput </script> |
16
Time Complexity: O(N)
Auxiliary Space: O(1)