the function is not returning any value

the problem is -- Given a list of ints, return True if the array contains a 3 next to a 3 somewhere.

Given a list of ints, return True if the array contains a 3 next to a 3 somewhere.

def has_33(list):
    for i in list:
        if (i==3 and (i+1)==3):
            return True
        else:
            pass
has_33([3, 3, 3])

i+1 is not the next element of the list, it adds 1 to the current element of the list.

You need to loop over the list indexes, not the values.

def has_33(list):
    for i in range(len(list)-1):
        if list[i] == 3 and list[i+1] == 3:
            return True
    return False

Or you can loop over pairs of elements using zip().

def has_33(list):
    for current, next in zip(list, list[1:]):
        if current == 3 == next:
            return True
    return False

def has_33(list):
    previous = None 
    for i in list:
        if previous is None:
            previous = i
        elif previous == 3 and i == 3:
            return True 
        else:
            previous = i
    return False 
print(has_33([1, 3, 3]))

Less clean than Barmar answer (not pythonic) but you could also store the previous value to compare it with the next one

You can store the current number in a variable so it compared with the next number in the next iteration:

def has_33(lst):
    last = None
    for i in lst:
        if i == last == 3:
            return True
        last = i
    return False

You can also use itertools.pairwise to iterate through the list by pairs of adjacent values:

from itertools import pairwise

def has_33(lst):
    return any(a == b == 3 for a, b in pairwise(lst))

An iterator solution, repeatedly finding the next 3 and checking the following element.

def has_33(lst):
    it = iter(lst)
    while 3 in it:
        if next(it, None) == 3:
            return True
    return False

You should use enumerate for next value and compare the current value

Code:

def has_33(ls):
    for i,v in enumerate(ls[:-1]):
        if v == 3 and ls[i+1] == 3:
            return True
    return False

The i goes from 1 to len(list)-1 so that we can compare the current and next values. After comparison, the list value is set to -1 just for reference.

def has_33(lst):
    for i in range(0, len(lst)-1):
        if (lst[i]==3 and lst[i+1]==3):
            return True
        else:
            pass
        lst[i]=-1
    return False 

l=[3,3,0]        
k=has_33(l)
print(k)