I got this DICE puzzle to solve and my mind is stuck on a scenario.
How many times did it occur in the trial, that exactly two 6 were rolled after each other? For example, in sequence 56611166626634416 it occurred twice, that exactly two 6 were thrown after each other.
Question is: how to avoid letting the counter count those 666.
Note: I have tried multiple trackers (keys), but then I have another issue which is:
IndexError: list index out of range
Throws=[6,6,2,6,6,6,3,6,6,3,6,6,6]
Counter_6 = 0
X=0
for i in range (0,len(Throws)):
if i==len(Throws) or i+1>len(Throws) or i+2>len(Throws):
key1= Throws[i]
key2=0
key3=0
elif i+2>=len(Throws):
key1 = Throws[i]
key2 = Throws[i + 1]
key3 = 0
else:
key1=Throws[i]
key2 = Throws[i + 1]
key3 = Throws[i + 2]
print("key 1 is", key1)
print("key 2 is", key2)
print("key 3 is", key3)
if key1==6 and key2==6 and key3!=6 and X==0:
Counter_6 = Counter_6 + 1
X=1
elif key1!=6 and key2 ==6 and key3==6 and X==0:
Counter_6 = Counter_6 + 1
X=1
elif key1==6 and key2==6 and key3==6:
Counter_6 = Counter_6
X=0
print("number of double 6 are: ",Counter_6)
Counter should be equal to 2
Here's easy and efficient solution without using any extending libraries. We define stack that is equal to amount of consecutive 6's we've seen, every time any digit other than 6 occur (or after looping over our throws) we check whether stack is equal to 2, if so we increment our counter_6 and reset stack to 0.
throws = [6,6,2,6,6,6,3,6,6,3,6,6,6]
counter_6 = 0
stack = 0
for i in throws:
if i == 6:
stack += 1
else:
if stack == 2:
counter_6 += 1
stack = 0
if stack == 2:
counter_6 += 1
print(counter_6) # --> 2