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 <bits/stdc++.h> 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
<script> // JavaScript code for the above approach let MAX = 128 // Function to check whether // a character is vowel or not function isVowel(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 function get(str, k) { let n = str.length; // Stores the count of // substring with K vowels let ans = 0; // Consider all substrings // beginning with str[i] for (let i = 0; i < n; i++) { let count = 0; // Consider all substrings // between [i, j] for (let 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 let s = "aeiou"; let K = 2; document.write(get(s, K)); // This code is contributed by Potta Lokesh </script> |
Output
4
Time Complexity: O(N2) where N is the length of the string.
Auxiliary Space: O(1)
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