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Count of K length substring containing at most X distinct vowels

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  • Last Updated : 12 Jan, 2022

Given string str of size N containing both uppercase and lowercase letters, and two integers K and X. The task is to find the count of substrings of size K containing at most X distinct vowels.

Examples:

Input: str = “TrueGoik”, K = 3, X = 2
Output: 6
Explanation: The five such substrings are “Tru”, “rue”, “ueG”, “eGo”, “Goi” and”oik”.

Input: str = “GeeksForGeeks”, K = 2, X = 2
Output: 12
Explanation: “Ge”, “ee”, “ek”, “ks”, “sf”, “fo”, “or”, “rG”, “Ge”, “ee”, “ek” and “ks”.

 

Approach: To solve the problem, first one have to generate all the substrings of length K. Then in each substring check if number of distinct vowels is less than X or not. Follow the steps mentioned below.

  • First generate all substrings of length K starting from each index i in [0, N – K].
  • Then for each substring of length K, do the following:
    • Keep a hash to store the occurrences of unique vowels.
    • Check if a new character in the substring is a vowel or not.
    • If it is a vowel, increment its occurrence in the hash and keep a count of distinct vowels found
    • Now for each substring, if the distinct count of vowels is less than or equal to X, increment the final count.
  • When all the substrings have been considered, print the final count.

Below is the implementation of the above approach:

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');
}
 
int getIndex(char ch)
{
    return (ch - 'A' > 26 ? ch - 'a' :
            ch - 'A');
}
 
// Function to find the count of K length
// substring with at most x distinct vowels
int get(string str, int k, int x)
{
    int n = str.length();
 
    // Initialize result
    int res = 0;
 
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i <= n - k; i++) {
        int dist_count = 0;
 
        // To store count of characters
        // from 'a' to 'z'
        vector<int> cnt(26, 0);
 
        // Consider all substrings
        // between str[i..j]
        for (int j = i; j < i + k; j++) {
 
            // If this is a new vowels
            // for this substring,
            // increment dist_count.
            if (isVowel(str[j])
                && cnt[getIndex(str[j])]
                == 0) {
                dist_count++;
            }
 
            // Increment count of
            // current character
            cnt[getIndex(str[j])]++;
        }
 
        // If count of distinct vowels
        // in current substring
        // of length k is less than
        // equal to x, then increment result.
        if (dist_count <= x)
            res++;
    }
 
    return res;
}
 
// Driver code
int main(void)
{
    string s = "TrueGoik";
    int K = 3, X = 2;
    cout << get(s, K, X);
    return 0;
}

Java




// Java code to implement above approach
import java.util.Arrays;
 
class GFG {
  static 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');
  }
 
  static int getIndex(char ch) {
    return (ch - 'A' > 26 ? ch - 'a' : ch - 'A');
  }
 
  // Function to find the count of K length
  // substring with at most x distinct vowels
  static int get(String str, int k, int x) {
    int n = str.length();
 
    // Initialize result
    int res = 0;
 
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i <= n - k; i++) {
      int dist_count = 0;
 
      // To store count of characters
      // from 'a' to 'z'
      int[] cnt = new int[26];
      Arrays.fill(cnt, 0);
 
      // Consider all substrings
      // between str[i..j]
      for (int j = i; j < i + k; j++) {
 
        // If this is a new vowels
        // for this substring,
        // increment dist_count.
        if (isVowel(str.charAt(j))
            && cnt[getIndex(str.charAt(j))] == 0) {
          dist_count++;
        }
 
        // Increment count of
        // current character
        cnt[getIndex(str.charAt(j))]++;
      }
 
      // If count of distinct vowels
      // in current substring
      // of length k is less than
      // equal to x, then increment result.
      if (dist_count <= x)
        res++;
    }
 
    return res;
  }
 
  // Driver code
  public static void main(String args[]) {
    String s = "TrueGoik";
    int K = 3, X = 2;
    System.out.println(get(s, K, X));
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3




# Python code for the above approach
 
# 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')
 
def getIndex(ch):
    return (ord(ch) - ord('a')) if (ord(ch) - ord('A')) > 26 else (ord(ch) - ord('A'))
 
# Function to find the count of K length
# substring with at most x distinct vowels
def get(str, k, x):
    n = len(str)
 
    # Initialize result
    res = 0
 
    # Consider all substrings
    # beginning with str[i]
    for i in range(n - k + 1):
        dist_count = 0
 
        # To store count of characters
        # from 'a' to 'z'
        cnt = [0] * 26
 
        # Consider all substrings
        # between str[i..j]
        for j in range(i, i + k):
 
            # If this is a new vowels
            # for this substring,
            # increment dist_count.
            if (isVowel(str[j]) and cnt[getIndex(str[j])] == 0):
                dist_count += 1
 
            # Increment count of
            # current character
            cnt[getIndex(str[j])] += 1
 
        # If count of distinct vowels
        # in current substring
        # of length k is less than
        # equal to x, then increment result.
        if (dist_count <= x):
            res += 1
 
    return res
 
# Driver code
 
s = "TrueGoik"
K = 3
X = 2
print(get(s, K, X))
 
# This code is contributed by Saurabh Jaiswal

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');
  }
 
  static int getIndex(char ch)
  {
    return (ch - 'A' > 26 ? ch - 'a' : ch - 'A');
  }
 
  // Function to find the count of K length
  // substring with at most x distinct vowels
  static int get(string str, int k, int x)
  {
    int n = str.Length;
 
    // Initialize result
    int res = 0;
 
    // Consider all substrings
    // beginning with str[i]
    for (int i = 0; i <= n - k; i++) {
      int dist_count = 0;
 
      // To store count of characters
      // from 'a' to 'z'
      int[] cnt = new int[26];
      // Arrays.fill(cnt, 0);
 
      // Consider all substrings
      // between str[i..j]
      for (int j = i; j < i + k; j++) {
 
        // If this is a new vowels
        // for this substring,
        // increment dist_count.
        if (isVowel(str[j])
            && cnt[getIndex(str[j])] == 0) {
          dist_count++;
        }
 
        // Increment count of
        // current character
        cnt[getIndex(str[j])]++;
      }
 
      // If count of distinct vowels
      // in current substring
      // of length k is less than
      // equal to x, then increment result.
      if (dist_count <= x)
        res++;
    }
 
    return res;
  }
 
  // Driver code
  public static void Main()
  {
    string s = "TrueGoik";
    int K = 3, X = 2;
    Console.WriteLine(get(s, K, X));
  }
}
 
// This code is contributed by ukasp.

Javascript




<script>
      // JavaScript code for the above approach
 
      // 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 getIndex(ch) {
          return ((ch.charCodeAt(0) - 'A'.charCodeAt(0)) > 26 ? (ch.charCodeAt(0) - 'a'.charCodeAt(0)) :
              (ch.charCodeAt(0) - 'A'.charCodeAt(0)));
      }
 
      // Function to find the count of K length
      // substring with at most x distinct vowels
      function get(str, k, x) {
          let n = str.length;
 
          // Initialize result
          let res = 0;
 
          // Consider all substrings
          // beginning with str[i]
          for (let i = 0; i <= n - k; i++) {
              let dist_count = 0;
 
              // To store count of characters
              // from 'a' to 'z'
              let cnt = new Array(26).fill(0);
 
              // Consider all substrings
              // between str[i..j]
              for (let j = i; j < i + k; j++) {
 
                  // If this is a new vowels
                  // for this substring,
                  // increment dist_count.
                  if (isVowel(str[j])
                      && cnt[getIndex(str[j])]
                      == 0) {
                      dist_count++;
                  }
 
                  // Increment count of
                  // current character
                  cnt[getIndex(str[j])]++;
              }
 
              // If count of distinct vowels
              // in current substring
              // of length k is less than
              // equal to x, then increment result.
              if (dist_count <= x)
                  res++;
          }
 
          return res;
      }
 
      // Driver code
 
      let s = "TrueGoik";
      let K = 3, X = 2;
      document.write(get(s, K, X));
 
// This code is contributed by Potta Lokesh
  </script>

 
 

Output
6

Time Complexity: O(N * K)
Auxiliary Space: O(N * K)

 


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