Question

Problem 1 - Find the Diphthongs

A diphthong is a composite vowel sound composed of two simple vowel sounds. We're going to be looking for these in words. For instance "ay" is normally a diphthong for instance in the words "lay" and "day", as is "ou" for instance in "loud," "lout."

We're going to simplify this by looking only for two vowels in a row in a word regardless of whether they are sounded out as diphthongs or not.  

Let me show some examples of the counting:

[Ae]rat[io]n; count = 2

F[ou]ndat[io]n; count = 2

Q[ue][ue]ing; count = 2

The reason that we are counting this as two, is because once we've used a vowel as part of a diphthong do not use it again.   

Note also that after the last pair, there's another vowel, but since the last e was used, the i doesn't pair, so it is not part of a diphthong.  

P[ae]an; count = 1

Several; count = 0

[Ae][ae][ae][ae][ae]; count = 5

This is obviously an invented word, but who cares, it's composed of five diphthong pairs.  

Glor[ia]; count = 1

For this exercise, the vowels are a, e, i, o, u, and y.  

Take in a word, output each diphthong pair and finally output the count.  

In terms of spaces, commas, and periods, let the words be divided, so don't count a dipthong over a space or other grammatical mark, meaning that for instance "a year" should group like "a [ye]ar" rather than "[a y][ea]r."  

Sample Output

linux1[150]% python3 find_diphthongs.py Enter a string with a lot of diphthongs: aeration ae io The diphthong count is 2 linux1[151]% python3 find_diphthongs.py Enter a string with a lot of diphthongs: aeitiour ae io The diphthong count is 2 linux1[152]% python3 find_diphthongs.py Enter a string with a lot of diphthongs: ayayay ay ay ay The diphthong count is 3 linux1[153]% python3 find_diphthongs.py Enter a string with a lot of diphthongs: argument The diphthong count is 0

If you want to eliminate a space between your letters and you're printing the characters separately, use:

print('a', 'e', sep='')

Answer 1

save as find_diphthongs.py

# Function that return true 
# if character ch is a vowel 
def isVowel(ch): 
    if ch in ['a', 'e', 'i', 'o', 'u','y']: 
        return True
    else:  
        return False

# Function to return the count of adjacent 
# vowel pairs in the given string 
def vowelPairs(s, n):   
    cnt = 0
    i=0
    
    while i<n-1:  
        step=1
        # If current character and the 
        # character after it are both vowels 
        if (isVowel(s[i]) and  isVowel(s[i + 1])): 
            print(s[i]+s[i+1])
            step = 2  #skip the used a vowel as part of a diphthong do not use it again
            cnt += 1  #count the diphthong
        i= i + step
          
    return cnt 
  
# Driver code 
def main():
    text=input("Enter a string with a lot of diphthongs: ")
    n = len(text) 
    c=vowelPairs(text, n)
    print("The diphthong count is ",c)

if __name__ == "__main__":
    main()