[SOLVED] Security problem...

Question

You are in charge of security at a casino, and there is a thief who is trying to steal the casino's money! Look over the security diagrams to make sure that you always have a guard between the thief and the money!
There is one money location, one thief, and any number of guards on each floor of the casino.

Task:
Evaluate a given floor of the casino to determine if there is a guard between the money and the thief, if there is not, you will sound an alarm.

My Solution:
n = input()
c = ""
for i in n:
if i != 'x':
c = c + i
a = c.index("G")
b = c.index("$")
c = c.index("T")
if b >= a and c >= b:
print("quiet")
elif a < b:
print("ALARM")
elif b < a and b < c:
print("quiet")
else:
print("ALARM")

#Unable to pass only one test case :(

Answer 1

#I think, this works..
#Observe the condition for guard just..

n = input()
c = ""
for i in n:
if i != 'x':
c = c + i
a=0
for i in c:
if i == 'G':
a = a + 1
#a = c.index("G")
b = c.index("$")
c = c.index("T")
if (b>a>c) or (b<a<c): #a between b and c(, guard between $ and T
print("quiet")
else:
print("ALARM")

#may this can be simplified in many ways in python.. but to understand problem, these steps clear that I hope.

**never give up in trying to solve anything..
hope it helps....

Answer 2

ANOTHER IDEA
REMOVE UNNECESSARY X AND CHECK IF THIEF IS BESIDE MONEY OR NOT
a=input().replace('x','')
if a.index('T') in [a.index('$')+1,a.index('$')-1]:
print('ALARM')
else:
print('quiet')

Answer 3

x = ''.join(input().split('x'))

y = ''.join(x.split('T$'))
z = ''.join(y.split('$T'))

print('ALARM'if '$' not in z else 'quiet')

Noob style: passes all tests.
#x removes space.
#y&z remove only the thieves with money
#If money is gone an alarm
#else - it's all quiet

Answer 4

I think this code works try it
If it does not work let me know

https://code.sololearn.com/c8l9l340NC6x

Keep learning & happy coding :D

Answer 5

I mean, For input G$T, it should output Alarm but your code , it's quiet.

Answer 6

$GT is failing case there...

Answer 7

Now only one case is failing...

Answer 8

Now it's failing the case G$T
Note: There may be more than one guard also..

Answer 9

That case is thr already :(
case 3 is failing for me

Answer 10

If possible cn u send the code pls
I tried all the ways...

Answer 11

Yo try this

setup = input()
setup = setup.replace('x', '')
S=setup.find('$')
T=setup.find('T')
if ((T - S) == 1) or ((T - S) ==-1):
print('ALARM')
else:
print('quiet')

Answer 12

cam= input().replace('x','')
x=cam.find('T')
y= cam.find('$')
if abs((x-y))==1:
print ('ALARM')
else :
print('quiet')

Answer 13

i = list(input())
x = []

for r in range(len(i)):
if i[r] != "x":
x.append(i[r])

z = "".join(x)

if "$T" in z or "T$" in z:
print("ALARM")
else:
print("quiet")

or:

i = input().replace("x", "")

if "$T" in i or "T$" in i:
print("ALARM")
else:
print("quiet")

Answer 14

#Swift #swift

var myString = readLine() ?? ")"
var th = 0
var mn = 0
var gt = 0
for i in myString{
if i=="$"{mn+=1}
else if i=="G"{mn=0; th=0}
else if i=="T"{th+=1}
if th>=1 && mn>=1{
print("ALARM"); gt+=1; break
}
};if gt == 0{print("quiet")}

Answer 15

a=input().replace("x","")

thief=a.index("T")
money=a.index("$")
range1=a[thief:money]
range2= a[money:thief]

if "G" in (range1 or range2):
print ("quiet")
else : print ("ALARM")