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+2

security challange

it get 3 right only import sys x=str(input()) T=0 G=0 z=0 T1=0 G1=0 n=len(x) S=x.index('$') try: G=x[0:S].index("G") except: print("ALARM") sys.exit() try: G1=x[S+1:n].index("G") except: print("ALARM") sys.exit() try: T=x[0:S].index("T") except: z=1 try: T1=x[S+1:n].index("T") except: z=2 if T>G: print("ALARM") elif T1<G1: print("ALARM") elif z==1 and z==2: print("quiet"),sys.exit() else: print("quiet")

2/26/2020 1:28:34 PM

abdosamyeid

4 Answers

New Answer

+1

As it's a little bit hard to follow your logic, I think you're not taking into account that there may be multiple Guards. Index returns only the first occurence.

+1

1. Firstly, try to get the index of thief and index of money from the given string. 2. Now iterate over the string from index with least value to index with highest value that you got in step 1 and search for any guard in between. If yes, then print quiet else ALARM.

0

If i remember good than "alert" should be only if between thief and money haven't any guard.. So in your case you in first step check if guard isn't before money then "alert" but there also maybe haven't any Thief Example: xxxx$xxxx There's no thief but also no guard. So in x.index('G') you get exception and print "alert".

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Alternatively, remove the 'x' and check if the index of money + or - 1 is the thief...