To Find Vowel-Substring From a String

You need to reset the count=0 at line number 17.

Try this code

def findSubstring(s, k):
    i = 0
    lst = []
    count = 0
    tempL = []
    
    while(i != len(s)):
        a = i+k
        temp = s[i:a]
        lst.append(temp)
        if a != len(s):
            i+=1
        else:
            break
    
    for word in lst:
        count = 0
        for alphabet in word:
            if alphabet in 'aeiou':
                count += 1
        tempL.append(count)
    print(lst)
    print(tempL)
    return 

s = 'azerdii'
k = 5
print(findSubstring(s, k))

Try the following:

def find_substr(s, k):
    substrings = [s[i:i+k] for i in range(len(s) - k + 1)] # list of substrings
    # vowels = [sum(c in 'aeiou' for c in s) for s in substrings]
    # print(vowels) # [2, 2, 3]
    return max(substrings, key=lambda s: sum(c in 'aeiou' for c in s))

print(find_substr('azerdii', 5)) # 'erdii'

If you un-comment the lines that are commented out, you will see the number of vowels is correctly computed as [2, 2, 3].

Here, sum(c in 'aeiou' for c in s) gets the number of vowels in a string s, which is equivalent to

count = 0
for alphabet in word:
    if alphabet in 'aeiou':
        count += 1

which in turn is the same as your code, except the line count = 0. After processing each word, you need to reset count. So try adding count = 0 in your code.

You could use a sliding window approach, for an optimal time complexity single-pass solution:

def find_substring_length_k_most_vowels(s: str, k: int) -> str:
    '''Returns first substring of length k that has the max number of vowels.'''
    vowels = set('aeiou')
    max_vowel_count = curr_vowel_count = 0
    max_window_start, max_window_end = 0, -1
    window_start = 0
    for window_end, ch in enumerate(s):
        if ch in vowels:
            curr_vowel_count += 1
        if window_end - window_start + 1 == k:
            if curr_vowel_count > max_vowel_count:
                max_vowel_count = curr_vowel_count
                max_window_start, max_window_end = window_start, window_end
            curr_vowel_count -= 1 if s[window_start] in vowels else 0
            window_start += 1
    return s[max_window_start:max_window_end + 1]


def main() -> None:
    s = 'azerdii'
    k = 5
    print(find_substring_length_k_most_vowels(s, k))


if __name__ == '__main__':
    main()

Output:

erdii

Brute Force Solution

Intuition :

1. Initialize a set and push all the elements of the sub string in that set(knowing that the element is a vowel). When its sizes == 5(as there are 5 vowels in English alphabet) then increase the counter by 1 Approach :

2. Make a function isVowel that return (1/true) when the passed character is a vowel otherwise (0/false).

3. Initialize 2 for loop a. Outer loop running from 0 to word.size()(till the end of the string). b. Inner loop running from i to word.size()(till the end of the string).

4. Now initialize a SET inside the outer for loop (as we don’t want that the values of previous iteration to persist in it).

5. In the inner for loop initialize a condition that if the J th Character value is not a vowel then break the loop.

6. And if it is a vowel then insert it in the initailzed SET

7. Now again initialize a condition that if the size of SET is == 5 then increase the counter by 1.

8. At last return the counter. Code:

   //  Function that checks that the given character is a vowel or not
   
    int isVowel(char c)
   {
   
    if ((c == 'a') || (c == 'e') || (c == 'i') || (c == 'o') || (c == 'u'))
   
   {
   
   
   return 1;
   
   }
   
   return 0;
   
   }
   
   
   int countVowelSubstrings(string word)
   
   {
   
      int counter = 0; // initalize the counter with 0
   
      for (int i = 0; i < word.size(); i++)
   
      {
   
   set<char> temp;// set initalized outside the for loop so as to avoid 
                   // the persitance of the pervious iteration values in 
                  //the set
   
   for (int j = i; j < word.size(); j++)
   
   {
   
       //  if the jth charcter is not a vowel then break the loop
   
       if (isVowel(word[j]) == 0)
   
       {
   
           break;
   
       }
   
       // else insert the char value in the SET
   
       temp.insert(word[j]);
   
       //  if size == 5 (as there are 5 vowels in English alphabet)  then 
       //  increase the counter by 1
   
   if (temp.size() == 5)
   
       {
   
           counter++;
   
       }
   
   }
   
       }
   
     return counter;
   
          }
        // Code By Ambuj Kumar
   

An almost one-liner functional approach:

from functools import reduce

def getVowelsLength(word, k):
    chunks = [word[i:i+k] for i in range(len(word)-(k-1))]
    return reduce(lambda x, y: x if x[1] > y[1] else y, list(zip(chunks, [sum(1 for l in w if l in 'aeiou') for w in chunks])))

which will give the correct answer: ('erdii', 3)