Why i am sticking here ? This is the casino problem, hard

Question

string = input( )

if "

Answer 1

Amit Ranjan Sahoo good job! But...

The problem tells you:

Input: string containing one "quot;, one "T", some "G" and some "x"

You need to

print("quiet")
if there is a G is in beetwen $ and T
else print("ALARM")

I challenge you to rewrite the sentence above in python in less than, let's be generous, 15 (readable) lines

If you have time, try doing it

Answer 2

Amit Ranjan Sahoo ,

i would suggest that you re-start a try with input() and getting the index from "G", "quot; and "T". (as you already have done)

with the 3 different elements we have, we could create 6 unique possible samples. but only 2 of them can give "quiet".

"quiet" is only possible, when "G" is in the MIDDLE of the others. so these two are: $GT and TG$.
these 2 are easy to check by the index you have already stored.
for these 2 output "quiet", and for all the others output "ALARM"

happy coding and good success!

Answer 3

Doug Warner ,

your post has no relation to the topic of this feed. this can be seen as spam.

> maybe you wanted to modify your post.

> otherwise it would be better if you start your own question, because the current discussion started one year ago. people may not be aware of your issue.

Answer 4

Come on! This should be python, you can even do it in less than 7 lines!

I know you can do it!

I'll give you a hint:
Take the substring from $ to T (you don't need to check if they're in string, the problem says they'll be there)

Answer 5

You do that so good. I challenge you to solve this problem with more $ and T , print "Alarm" if there is at least 1 $ between T and G else print "quiet"

Answer 6

Angelo
I did it but can't in 15 line but in 19 line of code 😀👍. Thank you for your challenge. I find the bug and did it in short❤️.
Here it is 👇👇👇

string = input()
m = 0
k = [ ]
if "G" in string :
gp = string.index("G")
for i in string :
if i in k :
gp = m
if i == "G" :
k.append(i)
m += 1
if gp < string.index("quot;) and gp > string.index("T") :
print("quiet")
elif gp>string.index("quot;) and gp<string.index("T") :
print("quiet")
else :
print("ALARM")
else :
print("ALARM")

Answer 7

Angelo
Bro yes,i found it in delay 😂,but i found 👍. Thnks for answering.

Answer 8

Great!

Answer 9

Lothar
I solved 👇
#number of floors
string = input( )

#here thif is one and place where the money is one.
#but number of guards are different, there can be one thif between two guards !

if "quot;not in string or "T"not in string or "G" not in string or "x"not in string :
if "G" in string and "quot; in string and "T"not in string :
print("quiet")
if "quot; not in string :
print("ALARM")
elif "G" not in string and "quot; in string and "T" in string and "x"in string :
print("ALARM")

if "quot; in string and "T" not in string and "G"not in string and "x" not in string :
print("ALARM")

if "quot; in string and "T" in string and "G" in string :
money_flor = string.index("quot;)
thif_flor = string.index("T")
guard_flor = string.index("G")

k = [ ]
m = 0
for i in string :
if i in k and "T" not in k :
guard_flor = m
k.append(i)
m += 1

for i in string :
if i == "G" and string.index(i)>money_flor :
guard_flor = string.index(i)
if money_flor < guard_flor and guard_flor < thif_flor :