Deja Vu : code coach peobel

Question

The given code will pass all the test in "Deja Vu" chellenge in coad coach expect the last one . I don't know why . Please help me out .

word = input()
def attention(data):

for i,p in zip(data[0:] , data[1:]):
if i is p:
print("Deja Vu")
break
else:
if p is data[-1]:
print("Unique")

attention(word)

Answer 1

Try to put this input:
Saran
It will come unique because your program will check only the consecutive letters
It should be Deja Vu

See a shorter method to solve the problem:

https://code.sololearn.com/c2mMDtrlEnoQ/?ref=app

Answer 2

here is my solution. slightly different approach

x = input("")

if len(x) > len(set(list(x))):
print("Deja Vu")
else:
print("Unique")

Answer 3

Print out the tupels, then you'll see, you miss a lot... Especially at the end of the strings.

Answer 4

Thanks Namit Jain
Finally I find the solution

word = input()
def atentuon(data):
string = []

for i in data:
if i in string:
return "Deja Vu"

else:
string.append(i)
return 'Unique'

def main():
print(atentuon(word))
main()

Answer 5

Namit Jain cuz I'm a new bee 😅
I'll get to learn more ...

Answer 6

Saran Ohh noo!
You are an awesome guy!
You tried!
And finally solved it! 🙃

Answer 7

Namit Jain Thanks a lot 😁

Answer 8

string = input()

u = False
for letter in range(0,len(string)):
unique = string[letter]

if unique in string[letter+1::]:
u = True
break
if u:
print("Deja Vu")
else:
print("Unique")

refer this code bro.

Answer 9

Saran great!
But why to make it lengthy? 🙃🤔

Answer 10

I'm in my noobie stages as well and I did it a little different the most. Idk if that's a good thing or not. So if someone has input, I'll gladly receive it.

rndm_char = input()
x = set(rndm_char)
re_strng = ''.join(x)

print('Unique' if len(rndm_char) == len(re_strng) else 'Deja Vu')