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Count of substrings containing exactly K distinct vowels

  • Last Updated : 29 Dec, 2021

Given string str of size N containing both uppercase and lowercase letters, and an integer K. The task is to find the count of substrings containing exactly K distinct vowels.

Examples:

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

Input: str = “TrueGoik”, K = 3
Output: 5
Explanation: The substrings are “TrueGo”, “rueGo”, “ueGo”, “eGoi” and “eGoik”.

 

Approach: The problem can be solved by generating all the substrings. From the generated substrings count the ones having K distinct vowels. Follow the steps mentioned below to implement the approach:

  • First generate all substrings starting from each index i in range [0, N]
  • Then for each substring, follow the steps:
    • Keep a hash array 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 K, increment the final count.
  • If for any loop to find substrings starting from i, the count of distinct vowels exceeds K, or, the substring length has reached string length, break the loop and look for substrings starting from i+1.
  • When all the substrings have been considered, print the final count.

Below is implementation of the above approach.

C++




// C++ program to count number of substrings
// with exactly k distinct vowels
#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 count number of substrings
// with exactly k unique vowels
int countkDist(string str, int k)
{
    int n = str.length();
  
    // Initialize result
    int res = 0;
  
    // Consider all substrings 
    // beginning with str[i]
    for (int i = 0; i < n; 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 < n; 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 distinct vowels count
            // becomes k then increment result
            if (dist_count == k) 
                res++;
  
            if (dist_count > k)
                break;
        }
    }
    return res;
}
  
// Driver code
int main()
{
    string str = "TrueGoik";
    int K = 3;
    cout << countkDist(str, K) << endl;
    return 0;
}

Java




// Java program to count number of substrings
// with exactly k distinct vowels
import java.util.*;
public 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');
}
  
static int getIndex(char ch)
{
    return (ch - 'A' > 26 ? ch - 'a'
            ch - 'A');
}
  
// Function to count number of substrings
// with exactly k unique vowels
static int countkDist(String str, int k)
{
    int n = str.length();
  
    // Initialize result
    int res = 0;
  
    // Consider all substrings 
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
        int dist_count = 0;
  
        // To store count of characters 
        // from 'a' to 'z'
        int cnt[] = new int[26];
        for(int t = 0; t < 26; t++) {
            cnt[t] = 0;
        }
          
        // Consider all substrings 
        // between str[i..j]
        for (int j = i; j < n; 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 distinct vowels count
            // becomes k then increment result
            if (dist_count == k) 
                res++;
  
            if (dist_count > k)
                break;
        }
    }
    return res;
}
  
// Driver code
public static void main(String args[])
{
    String str = "TrueGoik";
    int K = 3;
    System.out.println(countkDist(str, K));
}
}
  
// This code is contributed by Samim Hossain Mondal.

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 count number of substrings
# with exactly k unique vowels
def countkDist(str, k):
    n = len(str)
  
    # Initialize result
    res = 0
  
    # Consider all substrings
    # beginning with str[i]
    for i in range(n):
        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, n):
  
            # 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 distinct vowels count
            # becomes k then increment result
            if (dist_count == k):
                res += 1
  
            if (dist_count > k):
                break
    return res
  
# Driver code
s = "TrueGoik"
K = 3
  
print(countkDist(s, K))
  
# This code is contributed by Saurabh Jaiswal

C#




// C# program to count number of substrings
// with exactly k distinct vowels
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 count number of substrings
  // with exactly k unique vowels
  static int countkDist(string str, int k)
  {
    int n = str.Length;
  
    // Initialize result
    int res = 0;
  
    // Consider all substrings 
    // beginning with str[i]
    for (int i = 0; i < n; i++) {
      int dist_count = 0;
  
      // To store count of characters 
      // from 'a' to 'z'
      int []cnt = new int[26];
      for(int t = 0; t < 26; t++) {
        cnt[t] = 0;
      }
  
      // Consider all substrings 
      // between str[i..j]
      for (int j = i; j < n; 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 distinct vowels count
        // becomes k then increment result
        if (dist_count == k) 
          res++;
  
        if (dist_count > k)
          break;
      }
    }
    return res;
  }
  
  // Driver code
  public static void Main()
  {
    string str = "TrueGoik";
    int K = 3;
    Console.Write(countkDist(str, K));
  }
}
  
// This code is contributed by Samim Hossain Mondal.

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 count number of substrings
        // with exactly k unique vowels
        function countkDist(str, k)
        {
            let n = str.length;
  
            // Initialize result
            let res = 0;
  
            // Consider all substrings 
            // beginning with str[i]
            for (let i = 0; i < n; 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 < n; 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 distinct vowels count
                    // becomes k then increment result
                    if (dist_count == k)
                        res++;
  
                    if (dist_count > k)
                        break;
                }
            }
            return res;
        }
  
        // Driver code
        let s = "TrueGoik";
        let K = 3
  
        document.write(countkDist(s, K));
  
  // This code is contributed by Potta Lokesh
    </script>

 
 

Output
5

 

Time Complexity: O(N2)
Auxiliary Space: O(N2)

 


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