Question

Write a recursive function called abb pattern with a single parameter astr, which is a string. The function returns True if astr is a string of the form a nb 2n (n a-s followed by 2n b-s, where n is a positive integer) and False otherwise. For example, abb pattern("abb"), abb pattern("aabbbb"), and abb pattern("aaaabbbbbbbb") all return True, but abb pattern("") (parameter is an empty string), abb pattern("abbabb"), and abb pattern("aaabbbbb") all return False.

you may not use any built-in functions other than len, the index operator [] , the slice operator [:], and + (string or list concatenation). You also may not use any loops. Please note: The built-in function in is essentially a loop and may not be used in any of the functions in this question.

Answer 1

CODE:-

#take input from the user.
astr=input()
def abb_pattern(astr):
#the string must have atleast 3 characters to have abb pattern
#base condition if string is empty or doesnt have 'abb' then no abb pattern exist
if(len(astr)<3):
return False
#base condition the string sliced is abb then there is a abb pattern
if(len(astr)==3 and astr=='abb'):
return True
#the first character can never be 'b'
#so it returns false
if(astr[0]=='b'):
return False
#comes here if first character is a it comes here
#if last 2 characters are bb we slice first one character and last 2 characters
#[-2:] slices last 2 characters of the string
if(astr[-2:]=="bb"):
#recursive call that slices first 1 'a' and last 2 'bb'
return abb_pattern(astr[1:-2])
#if not in the above conditions comes here and that means the string is not in the pattern
return False
#printing the value of the function
print(abb_pattern(astr))

CODE ATTACHMENTS:-

OUTPUT:-

We use the base conditions to write the recursion.

We use slice[:] to minimise the string and send the recursive call.

We use astr[-2:] to get the last 2 characters.

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