Clash Royale CLAN TAG#URR8PPP

Print the length of the longest continuous sequence of alternating odd and even numbers

How would I go about reading integers until -1 is input and then printing the length of the longest continuous sequence of numbers where there are
alternating odd then even numbers?

I've achieved the first section but it went downhill from there.

Some testing lists:

[1,2,3,4,5,10,6,7,8,20,25,30,40,-1]
[6,7,8,20,25,30,40,1,2,3,4,5,10,15,20,-1]


Here is my code:

evenOdd=

while True:
try:
n=int(input())
if n != -1:
evenOdd.append(n)
except:
break
evenOdd=
longest = 0
length = 0
for i in range(len(evenOdd)):
if ((evenOdd[i-2]% 2 == 0) and (evenOdd[i-1]% 2 == 1) and (evenOdd[i]% 2 == 0):
length += 1
else:
longest = max(longest, length)
length = 0

print(longest)


2 Answers
2

One option would be to keep track of the longest sequence as you go:

longest =
current =

while True:
n = int(input("Enter value: "))
if n == -1:
break

if current and current[-1] % 2 != n % 2:
current.append(n)
else:
current = [n]

if len(current) > len(longest):
longest = current


The upside here is there's no post-processing to be done when the -1 is entered, the result is ready to go.

-1

You can apply itertools.groupby twice:

itertools.groupby

import itertools
d = [[1,2,3,4,5,10,6,7,8,20,25,30,40,-1], [6,7,8,20,25,30,40,1,2,3,4,5,10,15,20,-1]]
def key_func(d):
start= not d[0]%2
for i in d[1:]:
if i%2 == start:
start = (not i%2)
else:
return False
return True

for l in d:
new_l = [list(b) for _, b in itertools.groupby(l, key=lambda x:x%2)]
second_l = [[i for [i] in b] for a, b in itertools.groupby(new_l, key=lambda x:len(x) ==1) if a]
print(max(second_l, key=lambda x:[key_func(x), len(x)]))


Output:

[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5, 10, 15, 20, -1]


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