# How to check whether two lists are circularly identical in Python

For instance, I have lists:

```    a[0] = [1, 1, 1, 0, 0]
a[1] = [1, 1, 0, 0, 1]
a[2] = [0, 1, 1, 1, 0]
# and so on
```

They seem to be different, but if it is supposed that the start and the end are connected, then they are circularly identical.

The problem is, each list which I have has a length of 55 and contains only three ones and 52 zeros in it. Without circular condition, there are 26,235 (55 choose 3) lists. However, if the condition 'circular' exists, there are a huge number of circularly identical lists

Currently I check circularly identity by following:

```    def is_dup(a, b):
for i in range(len(a)):
if a == list(numpy.roll(b, i)): # shift b circularly by i
return True
return False
```

This function requires 55 cyclic shift operations at the worst case. And there are 26,235 lists to be compared with each other. In short, I need 55 26,235 (26,235 - 1) / 2 = 18,926,847,225 computations. It's about nearly 20 Giga!

Is there any good way to do it with less computations? Or any data types that supports circular?

First off, this can be done in `O(n)` in terms of the length of the list You can notice that if you will duplicate your list 2 times (`[1, 2, 3]`) will be `[1, 2, 3, 1, 2, 3]` then your new list will definitely hold all possible cyclic lists.

So all you need is to check whether the list you are searching is inside a 2 times of your starting list. In python you can achieve this in the following way (assuming that the lengths are the same).

```    list1 = [1, 1, 1, 0, 0]
list2 = [1, 1, 0, 0, 1]
print ' '.join(map(str, list2)) in ' '.join(map(str, list1 * 2))
```

Some explanation about my oneliner: `list * 2` will combine a list with itself, `map(str, [1, 2])` convert all numbers to string and `' '.join()` will convert array `['1', '2', '111']` into a string `'1 2 111'`.

As pointed by some people in the comments, oneliner can potentially give some false positives, so to cover all the possible edge cases:

```    def isCircular(arr1, arr2):
if len(arr1) != len(arr2):
return False

str1 = ' '.join(map(str, arr1))
str2 = ' '.join(map(str, arr2))
if len(str1) != len(str2):
return False

return str1 in str2 + ' ' + str2
```

P.S.1 when speaking about time complexity, it is worth noticing that `O(n)` will be achieved if substring can be found in `O(n)` time. It is not always so and depends on the implementation in your language (although potentially it can be done in linear time KMP for example).

P.S.2 for people who are afraid strings operation and due to this fact think that the answer is not good. What important is complexity and speed. This algorithm potentially runs in `O(n)` time and `O(n)` space which makes it much better than anything in `O(n^2)` domain. To see this by yourself, you can run a small benchmark (creates a random list pops the first element and appends it to the end thus creating a cyclic list. You are free to do your own manipulations)

```    from random import random
bigList = [int(1000 * random()) for i in xrange(10**6)]
bigList2 = bigList[:]
bigList2.append(bigList2.pop(0))

# then test how much time will it take to come up with an answer
from datetime import datetime
startTime = datetime.now()
print isCircular(bigList, bigList2)
print datetime.now() - startTime    # please fill free to use timeit, but it will give similar results
```

0.3 seconds on my machine. Not really long. Now try to compare this with `O(n^2)` solutions. While it is comparing it, you can travel from US to Australia (most probably by a cruise ship)

From: stackoverflow.com/q/26924836

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