Find all substrings containing exactly K unique vowels

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

Examples:

Input:  str = “aeiou”, K = 2
Output: “ae”, “ei”, “io”, “ou”
Explanation: These are the substrings containing exactly 2 distinct vowels.

Input: str = “TrueGeek”, K = 3
Output: “”
Explanation: Though the string has more than 3 vowels but there is not three unique vowels.
So the answer is empty.

Input: str = “TrueGoik”, K = 3
Output: “TrueGo”, “rueGo”, “ueGo”, “eGoi”, “eGoik”

Approach: This problem can be solved by greedy approach. Generate the substrings and check for every substring. Follow the steps mentioned below to solve the problem:

• First generate all substrings starting from each index i in range 0 to N.
• Then for each substring:
  • 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, print the substring.
• 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.

Below is the implementation of the above approach.

C++

// C++ program to find all 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 find all substrings
// with exactly k unique vowels
void countkDist(string str, int k)
{
    int n = str.length();

    // 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 vowel
            // 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 print the
            // substring.
            if (dist_count == k) {
                cout << str.substr(i, j - i + 1) << endl;
            }

            if (dist_count > k)
                break;
        }
    }
}

// Driver code
int main()
{
    string str = "TrueGoik";
    int K = 3;
    countkDist(str, K);
    return 0;
}

Java

// Java program to find all subStrings with
// exactly k distinct vowels
import java.util.*;

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');
  }

  static int getIndex(char ch)
  {
    return (ch - 'A' > 26 ?
            ch - 'a' : ch - 'A');
  }

  // Function to find all subStrings
  // with exactly k unique vowels
  static void countkDist(String str, int k)
  {
    int n = str.length();

    // 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];

      // Consider all subStrings
      // between str[i..j]
      for (int j = i; j < n; j++) {
        String print  = new String(str);
        // If this is a new vowel
        // 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 print the
        // subString.
        if (dist_count == k) {
          System.out.print(print.substring(i, j +1) +"\n");
        }

        if (dist_count > k)
          break;
      }
    }
  }

  // Driver code
  public static void main(String[] args)
  {
    String str = "TrueGoik";
    int K = 3;
    countkDist(str, K);
  }
}

// This code is contributed by Rajput-Ji

Python3

# python3 program to find all substrings with
# exactly k distinct vowels
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')

def getIndex(ch):
    if ord(ch) - ord('A') > 26:
        return ord(ch) - ord('a')
    else:
        return ord(ch) - ord('A')

# Function to find all substrings
# with exactly k unique vowels
def countkDist(str, k):
    n = len(str)

    # Consider all substrings
    # beginning with str[i]
    for i in range(0, n):
        dist_count = 0

        # To store count of characters
        # from 'a' to 'z'
        cnt = [0 for _ in range(26)]

        # Consider all substrings
        # between str[i..j]
        for j in range(i, n):

            # If this is a new vowel
            # 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 print the
            # substring.
            if (dist_count == k):
                print(str[i:i+j - i + 1])

            if (dist_count > k):
                break

# Driver code
if __name__ == "__main__":

    str = "TrueGoik"
    K = 3
    countkDist(str, K)

    # This code is contributed by rakeshsahni

C#

// C# program to find all 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 find all substrings
  // with exactly k unique vowels
  static void countkDist(string str, int k)
  {
    int n = str.Length;

    // 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];

      // Consider all substrings
      // between str[i..j]
      for (int j = i; j < n; j++) {

        // If this is a new vowel
        // 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 print the
        // substring.
        if (dist_count == k) {
          Console.WriteLine(
            str.Substring(i, j - i + 1));
        }

        if (dist_count > k)
          break;
      }
    }
  }

  // Driver code
  public static void Main()
  {
    string str = "TrueGoik";
    int K = 3;
    countkDist(str, K);
  }
}

// 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 all substrings
      // with exactly k unique vowels
      function countkDist(str, k) {
          let n = str.length;

          // 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 vowel
                  // 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 print the
                  // substring.
                  if (dist_count == k) {
                      document.write(str.slice(i, i + j - i + 1) + '<br>');
                  }

                  if (dist_count > k)
                      break;
              }
          }
      }

      // Driver code

      let s = "TrueGoik";
      let K = 3

      countkDist(s, K);

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

Output

TrueGo
rueGo
ueGo
eGoi
eGoik

Time Complexity: O(N²)
Auxiliary Space: O(N²) to store the resulting substrings