Count of K length substrings containing exactly X vowels

Given string str of N characters containing both uppercase and lowercase English alphabets, the task is to find the count of substrings of size K containing exactly X vowels.

Examples:

Input: str = “GeeksForGeeks”, K = 2, X = 2
Output: 2
Explanation: The given string only contains 2 substrings of size 2 consisting of 2 vowels. They are “ee” and “ee”.

Input: str = “TrueGeek”, K = 3, X = 2
Output: 5

Approach: The given problem can be solved using a sliding window technique by maintaining a window of size K and keeping track of the number of vowels encountered in the current window. If the count of vowels in the current window is equal to X, increment the final required count by 1.

Below is the implementation of the above approach:

C++

// C++ program of the 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
// K-sized substring having X vowels
int cntSubstr(string str, int K, int X)
{
    // Stores the number of vowels
    // in the current window
    int vow = 0;

    for (int i = 0; i < K; i++)
        if (isVowel(str[i]))
            vow++;

    // Stores the count of K length
    // substring with X vowels
    int ans = vow == X ? 1 : 0;

    for (int i = 1; i < str.length(); i++) {

        // Remove (i - 1)th character
        // from the current window
        vow = isVowel(str[i - 1]) ? vow - 1 : vow;

        // Insert (i - 1 + K)th character
        // from the current window
        vow = isVowel(str[i - 1 + K]) ? vow + 1 : vow;

        if (vow == X)

            // Increment answer
            ans++;
    }

    // Return Answer
    return ans;
}

// Driver code
int main(void)
{
    string s = "TrueGeek";
    int K = 3, X = 2;

    cout << cntSubstr(s, K, X);

    return 0;
}

Java

// Java code to implement the above approach
import java.io.*;
class GFG
{

  // 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
  // K-sized subString having X vowels
  static int cntSubstr(String str, int K, int X)
  {

    // Stores the number of vowels
    // in the current window
    int vow = 0;

    for (int i = 0; i < K; i++)
      if (isVowel(str.charAt(i)))
        vow++;

    // Stores the count of K length()
    // subString with X vowels
    int ans = vow == X ? 1 : 0;

    for (int i = 1; i < str.length(); i++) {

      // Remove (i - 1)th character
      // from the current window
      vow = isVowel(str.charAt(i - 1)) ? vow - 1 : vow;

      // Insert (i - 1 + K)th character
      // from the current window
      if(i - 1 +  K < str.length())
        vow = isVowel(str.charAt(i - 1 + K)) ? vow + 1 : vow;

      if (vow == X)

        // Increment answer
        ans++;
    }

    // Return Answer
    return ans;
  }

  // Driver code
  public static void main (String[] args) {
    String s = "TrueGeek";
    int K = 3, X = 2;

    System.out.println(cntSubstr(s, K, X));
  }
}

// This code is contributed by Shubham Singh

Python3

# Python3 program of the 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
# K-sized substring having X vowels
def cntSubstr(str, K, X):

    # Stores the number of vowels
    # in the current window
    vow = 0

    for i in range(0, K):
        if (isVowel(str[i])):
            vow += 1

    # Stores the count of K length
    # substring with X vowels
    ans = 1 if vow == X else 0

    for i in range(1, len(str) - K + 1):

        # Remove (i - 1)th character
        # from the current window
        vow = vow - 1 if isVowel(str[i - 1]) else vow

        # Insert (i - 1 + K)th character
        # from the current window
        vow = vow + 1 if isVowel(str[i - 1 + K]) else vow

        if (vow == X):

            # Increment answer
            ans += 1

    # Return Answer
    return ans

# Driver code
if __name__ == "__main__":

    s = "TrueGeek"
    K, X = 3, 2

    print(cntSubstr(s, K, X))

# This code is contributed by rakeshsahni

C#

// C# code to implement the 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
  // K-sized substring having X vowels
  static int cntSubstr(string str, int K, int X)
  {
    // Stores the number of vowels
    // in the current window
    int vow = 0;

    for (int i = 0; i < K; i++)
      if (isVowel(str[i]))
        vow++;

    // Stores the count of K length
    // substring with X vowels
    int ans = vow == X ? 1 : 0;

    for (int i = 1; i < str.Length; i++) {

      // Remove (i - 1)th character
      // from the current window
      vow = isVowel(str[i - 1]) ? vow - 1 : vow;

      // Insert (i - 1 + K)th character
      // from the current window
      if(i - 1 +  K < str.Length)
        vow = isVowel(str[i - 1 + K]) ? vow + 1 : vow;

      if (vow == X)

        // Increment answer
        ans++;
    }

    // Return Answer
    return ans;
  }

  // Driver code
  public static void Main()
  {
    string s = "TrueGeek";
    int K = 3, X = 2;

    Console.Write(cntSubstr(s, K, X));
  }
}

// This code is contributed by Samim Hossain Mondal.

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
        // K-sized substring having X vowels
        function cntSubstr(str, K, X)
        {

            // Stores the number of vowels
            // in the current window
            let vow = 0;

            for (let i = 0; i < K; i++)
                if (isVowel(str[i]))
                    vow++;

            // Stores the count of K length
            // substring with X vowels
            let ans = vow == X ? 1 : 0;
            for (let i = 1; i < str.length; i++) {

                // Remove (i - 1)th character
                // from the current window
                vow = isVowel(str[i - 1]) ? vow - 1 : vow;

                // Insert (i - 1 + K)th character
                // from the current window
                vow = isVowel(str[i - 1 + K]) ? vow + 1 : vow;

                if (vow == X)

                    // Increment answer
                    ans++;
            }

            // Return Answer
            return ans;
        }

        // Driver code
        let s = "TrueGeek";
        let K = 3, X = 2;

        document.write(cntSubstr(s, K, X));

  // This code is contributed by Potta Lokesh
    </script>

Output

5

Time Complexity: O(N)
Auxiliary space: O(1)