# Count of substrings containing exactly K vowels

Given string str containing both uppercase and lowercase letters, and an integer K. The task is to find the count of substrings containing exactly K vowels (maybe repetitive).

Examples:

Input: str = “aeiou”, K = 2
Output: 4
Explanation: The substrings are “ae”, “ei”, “io”, “ou”.

Input: str = “TrueGeek”, K = 3
Output: 5
Explanation: All such possible substrings are:
“TrueGe”, “rueGe”, “ueGe”, “eGee”, “eGeek”.

Approach: To solution of this problem is based on greedy approach. Generate all substrings and for each substring check the count of vowels. Follow the steps mentioned below.

• Generate all the substrings. For each substring do the following
• Store the count of occurrences of vowels.
• Check if a new character in the substring is a vowel or not.
• If it is a vowel, increment count of vowels found
• Now for each substring, if the count of vowels is K, increment the final count.
• Return the final count as the required answer at the end.

Below is the implementation of the above code.

## C++

 `// C++ code to implement above approach ` `#include ` `using` `namespace` `std; ` ` `  `#define MAX 128 ` ` `  `// Function to check whether ` `// a character is vowel or not ` `bool` `isVowel(``char` `x) ` `{ ` `    ``return` `(x == ``'a'` `|| x == ``'e'` `||  ` `            ``x == ``'i'` `|| x == ``'o'` `|| x == ``'u'`  `            ``|| x == ``'A'` `|| x == ``'E'` `||  ` `            ``x == ``'I'` `|| x == ``'O'` `|| x == ``'U'``); ` `} ` ` `  `// Function to find the count of ` `// substring with k vowels ` `int` `get(string str, ``int` `k) ` `{ ` ` `  `    ``int` `n = str.length(); ` ` `  `    ``// Stores the count of ` `    ``// substring with K vowels ` `    ``int` `ans = 0; ` ` `  `    ``// Consider all substrings  ` `    ``// beginning with str[i] ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``int` `count = 0; ` ` `  `        ``// Consider all substrings  ` `        ``// between [i, j] ` `        ``for` `(``int` `j = i; j < n; j++) { ` ` `  `            ``// If this is a vowel, for this ` `            ``// substring, increment count. ` `            ``if` `(isVowel(str[j])) { ` `                ``count++; ` `            ``} ` ` `  `            ``// If vowel count becomes k, ` `            ``// then increment final count. ` `            ``if` `(count == k) { ` `                ``ans++; ` `            ``} ` ` `  `            ``if` `(count > k) ` `                ``break``; ` `        ``} ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main(``void``) ` `{ ` `    ``string s = ``"aeiou"``; ` `    ``int` `K = 2; ` `    ``cout << get(s, K); ` `    ``return` `0; ` `}`

## Java

 `// Java code to implement above approach ` `class` `GFG ` `{ ` ` `  `  ``static` `final` `int` `MAX = ``128``; ` ` `  `  ``// Function to check whether ` `  ``// a character is vowel or not ` `  ``static` `boolean` `isVowel(``char` `x) ` `  ``{ ` `    ``return` `(x == ``'a'` `|| x == ``'e'` `||  ` `            ``x == ``'i'` `|| x == ``'o'` `|| x == ``'u'`  `            ``|| x == ``'A'` `|| x == ``'E'` `||  ` `            ``x == ``'I'` `|| x == ``'O'` `|| x == ``'U'``); ` `  ``} ` ` `  `  ``// Function to find the count of ` `  ``// subString with k vowels ` `  ``static` `int` `get(String str, ``int` `k) ` `  ``{ ` ` `  `    ``int` `n = str.length(); ` ` `  `    ``// Stores the count of ` `    ``// subString with K vowels ` `    ``int` `ans = ``0``; ` ` `  `    ``// Consider all subStrings  ` `    ``// beginning with str[i] ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `      ``int` `count = ``0``; ` ` `  `      ``// Consider all subStrings  ` `      ``// between [i, j] ` `      ``for` `(``int` `j = i; j < n; j++) { ` ` `  `        ``// If this is a vowel, for this ` `        ``// subString, increment count. ` `        ``if` `(isVowel(str.charAt(j))) { ` `          ``count++; ` `        ``} ` ` `  `        ``// If vowel count becomes k, ` `        ``// then increment final count. ` `        ``if` `(count == k) { ` `          ``ans++; ` `        ``} ` ` `  `        ``if` `(count > k) ` `          ``break``; ` `      ``} ` `    ``} ` `    ``return` `ans; ` `  ``} ` ` `  `  ``// Driver code ` `  ``public` `static` `void` `main(String[] args) ` `  ``{ ` `    ``String s = ``"aeiou"``; ` `    ``int` `K = ``2``; ` `    ``System.out.print(get(s, K)); ` `  ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar`

## Python3

 `# python code to implement above approach ` `MAX` `=` `128` ` `  `# Function to check whether ` `# a character is vowel or not ` `def` `isVowel(x): ` ` `  `    ``return` `(x ``=``=` `'a'` `or` `x ``=``=` `'e'` `or` `            ``x ``=``=` `'i'` `or` `x ``=``=` `'o'` `or` `x ``=``=` `'u'` `            ``or` `x ``=``=` `'A'` `or` `x ``=``=` `'E'` `or` `            ``x ``=``=` `'I'` `or` `x ``=``=` `'O'` `or` `x ``=``=` `'U'``) ` ` `  `# Function to find the count of ` `# substring with k vowels ` `def` `get(``str``, k): ` ` `  `    ``n ``=` `len``(``str``) ` ` `  `    ``# Stores the count of ` `    ``# substring with K vowels ` `    ``ans ``=` `0` ` `  `    ``# Consider all substrings ` `    ``# beginning with str[i] ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``count ``=` `0` ` `  `        ``# Consider all substrings ` `        ``# between [i, j] ` `        ``for` `j ``in` `range``(i, n): ` ` `  `            ``# If this is a vowel, for this ` `            ``# substring, increment count. ` `            ``if` `(isVowel(``str``[j])): ` `                ``count ``+``=` `1` ` `  `            ``# If vowel count becomes k, ` `            ``# then increment final count. ` `            ``if` `(count ``=``=` `k): ` `                ``ans ``+``=` `1` ` `  `            ``if` `(count > k): ` `                ``break` ` `  `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``s ``=` `"aeiou"` `    ``K ``=` `2` `    ``print``(get(s, K)) ` ` `  `# This code is contributed by rakeshsahni `

## C#

 `// C# code to implement above approach ` `using` `System; ` `class` `GFG { ` ` `  `  ``// Function to check whether ` `  ``// a character is vowel or not ` `  ``static` `bool` `isVowel(``char` `x) ` `  ``{ ` `    ``return` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| x == ``'o'` `            ``|| x == ``'u'` `|| x == ``'A'` `|| x == ``'E'` `            ``|| x == ``'I'` `|| x == ``'O'` `|| x == ``'U'``); ` `  ``} ` ` `  `  ``// Function to find the count of ` `  ``// substring with k vowels ` `  ``static` `int` `get``(``string` `str, ``int` `k) ` `  ``{ ` ` `  `    ``int` `n = str.Length; ` ` `  `    ``// Stores the count of ` `    ``// substring with K vowels ` `    ``int` `ans = 0; ` ` `  `    ``// Consider all substrings ` `    ``// beginning with str[i] ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `      ``int` `count = 0; ` ` `  `      ``// Consider all substrings ` `      ``// between [i, j] ` `      ``for` `(``int` `j = i; j < n; j++) { ` ` `  `        ``// If this is a vowel, for this ` `        ``// substring, increment count. ` `        ``if` `(isVowel(str[j])) { ` `          ``count++; ` `        ``} ` ` `  `        ``// If vowel count becomes k, ` `        ``// then increment final count. ` `        ``if` `(count == k) { ` `          ``ans++; ` `        ``} ` ` `  `        ``if` `(count > k) ` `          ``break``; ` `      ``} ` `    ``} ` `    ``return` `ans; ` `  ``} ` ` `  `  ``// Driver code ` `  ``public` `static` `void` `Main() ` `  ``{ ` `    ``string` `s = ``"aeiou"``; ` `    ``int` `K = 2; ` `    ``Console.WriteLine(``get``(s, K)); ` `  ``} ` `} ` ` `  `// This code is contributed by ukasp.`

## Javascript

 ``

Output

`4`

Time Complexity: O(N2) where N is the length of the string.
Auxiliary Space: O(1)

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