Python - Roadrunner - test case 3(hidden) fails - what am i missing?

Question

dist_apart = 50

dist_to_safety = int(input())

rr_speed = int(input())

k_speed = int(input())

catch_rate = k_speed - rr_speed

catch_time = dist_apart/catch_rate

rr_safe_time = dist_to_safety/rr_speed

if rr_safe_time > catch_time:

print("Yum")

elif rr_safe_time < catch_time:
print("Meep Meep")

Answer 1

Your catch time is wrong. It is as if you considered that the road runner is not moving. The distance is changing with time, you forgot the add that as well.

Here is a fix:
Instead of trying to figure out catch_time, consider it as a race. If the coyote arrives first, he wins. Else, the road runner wins.

Which means, catch_time will now represent the time needed for the coyote to go to safe zone. Thus this formula:

catch_time = (dist_apart+dist_to_safety) / k_speed

Also, catch_rate will be useless if you look at the situation like I did, so you can remoce it.

Answer 2

Adam technically, the road runner will be safe.

If they voth arrive to safe zone same time, then the road runner will already be safe.

But what matters is that you deliberately left it out, so no problem 👌

Answer 3

Thanks Aymane. The code below now passes all test cases.

dist_apart = 50

dist_to_safety = int(input())

rr_speed = int(input())

k_speed = int(input())

k_finish_time = (dist_apart + dist_to_safety)/k_speed

rr_finish_time = dist_to_safety/rr_speed

if rr_finish_time > k_finish_time :

print("Yum")

elif rr_finish_time < k_finish_time :

print("Meep Meep")

Answer 4

Adam even tho it is now fixed, you forgot about one test case:

rr_finish_time == k_finish_time

But nice code, and good choice of variables 😁👍

Answer 5

distance_safety = int(input())
roadrunners_speed = int(input())
coyotes_speed = int(input())

if distance_safety / roadrunners_speed < (distance_safety + 50) / coyotes_speed:
print("Meep Meep")
else:
print("Yum")

Answer 6

Thanks Aymane.

I deliberately left out that test case as the question doesn't specify what happens in the event of a draw.

Answer 7

Had the same problem with test case 3... Made modifications as suggested and all cases worked but I don't get why using closing speed didn't work.... In the first code above closing time was calculated using closing rate, not just the coyote speed... In my head using the rate here accounts for both parties moving...
Speed 1 = 20
Speed 2 = 5
Closing rate = 15
So every second the become 15 feet closer....
I must be missing something!!

Answer 8

I compare times needed to beat distances to safe place. If coyote beat this distance faster, that mean he coughts roadrunner.

Answer 9

s = int(input())
r = int(input())
c = int(input())
d = 50

if (s + d) / c < s / r:
print("Yum")
else:
print("Meep Meep")

Answer 10

For those who cant do math like me, just replecate a real race🏁, it works for me

My roadrunner code

rrposis = int(input())
rrspeed = int(input())
caospeed = int(input())
caoposis = rrposis + 50
while caoposis > rrposis and rrposis>0:
rrposis -= rrspeed
caoposis -= caospeed
print(('Meep Meep','Yum')[rrposis>0])

The last print statment
Is just a fancy if statment trick, basicly im printing yum if rrposis >0

Answer 11

distancia = int(input())
roadrunner = int(input())
coyote =int(input())

coyote_t = 50/coyote
roadrunner_t = distancia/roadrunner

x= coyote_t - roadrunner_t

if x>0:
print('Meep Meep')
else:
print('Yum')

Answer 12

distance = 50
distance_safe = int(input())
road_runner_speed = int(input())
coyote_speed = int(input())
time = distance_safe / road_runner_speed
if time * coyote_speed <= distance:
print('Meep Meep')
else:
print('Yum')

Answer 13

d_safety= int(input())
r= int(input())
c= int(input())
time_c= (50+r)//c
time_r = d_safety//r
if time_r < time_c :
print("Meep Meep")
else:
print("Yum")

Answer 14

In test 3 the roadrunner's speed is greater than the coyote's speed. Thus catch_rate becomes negative and the program breaks. This fixes it:

if rr_safe_time > catch_time and catch_time > 0