Trouble with Has22 problem.

Question

Problem:
Given an array of ints, return True if the array contains a 2 next to a 2 somewhere.

E.g:
has22([1, 2, 2]) → True
has22([1, 2, 1, 2]) → False
has22([2, 1, 2]) → False

My attempt: (Doesn't pass all conditions)
https://code.sololearn.com/clKEVJ4Tgu11/?ref=app
What could be the mistake?

Also I believe that python has a potential to do this task in one line?
So anyone expert in oneliners?

Answer 1

# Created by IQ >= 80; >> False

lst1 = [3,5,5,6]
lst2 = [3,2,2,7]

def func(nums):
for i in range(len(nums)-1):
if nums[i] ==2:
if nums[i] == nums[i+1]:
return True
return False

print(func(lst1))
print(func(lst2))

Answer 2

It is wrong solution. Your func return True if it will receive value equal to 2. but you need to find two next to each other. Use indexes of value 2.

Answer 3

Or iter list and check for element[i] == element[i+1] == 2
And find a way to avoid indexerror

Answer 4

We find only num to or any mun next to each other?

Answer 5

def fun(a): return True if a[:2] == [2, 2] else fun(a[1:]) if a[2:] else False

print(fun(a))

Answer 6

lst1 = [3,2,7,6]
lst2 = [3,2,2,7]

def func(nums):
for i in range(len(nums)-1):
if nums[i] == nums[i+1]:
return True
return False

print(func(lst1))
print(func(lst2))

Answer 7

My english not so good. I not understand problem at full.

Answer 8

IQ >= 80; >> False No one. It simple and have a lot different solution. Good luck to learn!

Answer 9

IQ >= 80; >> False
This is close, but not quite what you need.
I still striggle with one-liners

[print(lst[i]==lst[i+1]) if lst[i]==2 else '' for i in range(len(lst)-1)]

Answer 10

IQ >= 80; >> False
Let me know if this works

[print('true') if (lst[i]==2 and lst[i+1]==2) else '' for i in range(len(lst)-1)]

Answer 11

Rik Wittkopp Cool. But somehow can you convert it to return statement. Because the test support only function def

Answer 12

IQ >= 80; >> False
I will have a think.
🤔

Answer 13

IQ >= 80; >> False
I can get the one-liner to work with the def, & I changed the output to reflect True or False.
Is it a neccessity to use return?

Try this instead

def twos(x):
[print (True) if (x[i]==2 and x[i+1]==2) else False for i in range(len(x)-1)]

(twos(lst))

Answer 14

Rik Wittkopp website doesn't allow printing.
https://codingbat.com/prob/p119308

Answer 15

Valmob101 Thanks buddy, you rock!!

Answer 16

def func(nums):
i = 0
while i+1 < len(nums):
if nums[i] == nums[i+1] == 2:
return True
break
else:
i+=1
return False

print(func([4,1,2,2]))

Answer 17

Rik Wittkopp Thanks for the solution. I think this passed 90% of the condition. But it doesn't passed some hidden conditions.

Answer 18

Rik Wittkopp How to include only 2. This types of arrays didn't passed: [1,5,5,3]

Answer 19

Rik Wittkopp thanks, now it works.
But how much brain cells do we need to do this in one line?