Code Coach - It's a sign (Python)

Question

Hello!
I found out how to reverse the string but I don't know how to output "trash" if the while loop runs without output...is there a way for that?

inp=input()
words=inp.split()

for i in words:
x=len(i)
back=""
while x >= 0:
back+=i[x-1]
x-=1
if i == back:
print("Open")
else:
continue

Answer 1

You can add a counter, if the counter remains zero by the end, then execute the print "trash" statement.

Answer 2

Your code has a problem right at the beginning, you are not accepting 4 inputs to test, only one.

You will need a loop to generate 4 words, and store them in a list.

Prove that first, then build a bit of code to test each word.

Answer 3

Áron Szűcs ,

to check if a word is a palindrome (reading the same forward and backward) we can reverse the word and compare it to the not reversed word:

...
if word == word[::-1]:
# word is palindrome
...

[::-1] is a slice that reverses strings and some other container types.

Answer 4

Áron Szűcs

You are getting closer. 👍

Your code now prints "Open" every time it sees a palindrome, which means that if you have more than one palindrome in a box, then your output will not meet task requirements

Answer 5

I tried it and solves 3 out of the 5 test cases... What can be the problem with the other 2?

inp=input()
words=inp.split()
count=0
for i in words:
x=len(i)
back=""
while x >= 0:
back+=i[x-1]
x-=1
if i == back:
print("Open")
count+=1
else:
continue
if count == 0:
print("Trash")

Answer 6

Yeah right I forgot the input is always 4 different words... This code works in 4/5 test cases, what could be the trouble with it?

word1=input()
word2=input()
word3=input()
word4=input()

inp=[word1,word2,word3,word4]

count=0
for i in inp:
x=len(i)
back=""
while x >= 0:
back+=i[x-1]
x-=1
if i == back:
print("Open")
count+=1
else:
continue
if count == 0:
print("Trash")

Answer 7

I managed to solve it with the counter... Thanks to both of y'all! :)