Find if the array contain a 2 next to a 2

Question

I am stuck on this problem

Given an array of ints, return True if the array contains a 2 next to a 2 somewhere.

has22([1, 2, 2]) → True
has22([1, 2, 1, 2]) → False
has22([2, 1, 2]) → False

I know the basic idea (there are syntax errors) but I can't implement it. I would also like to know what type of problem this is, eg. graph, search?

def has22(nums):
for x in nums:
    if ( (nums[x] = 2) and (nums[x+1] = 2) )
        return True

return False 

Solution 1

def has22(nums):
    return any(x == y == 2 for x, y in zip(nums, nums[1:]))

>>> has22([1, 2, 2])
True
>>> has22([1, 2, 1, 2])
False
>>> has22([2, 1, 2])
False

In Python 2 use: from itertools import izip if you want a lazy zip

Solution 2

def has22(nums):
    for x in range(len(nums)-1):
        if (nums[x] == 2) and (nums[x+1] == 2):
            return True
    return False

I have just corrected your code. It runs in linear time so don't see any reason to work on it further.

Here is the running code on codebunk. http://codebunk.com/bunk#-Ivk7Xw2blX3cIWavI17

Solution 3

def has22(nums):
    it = iter(nums)
    return any(x == 2 == next(it, None) for x in it)

>>> has22([1, 2, 2])
True
>>> has22([1, 2, 1, 2])
False
>>> has22([2, 1, 2])
False

Solution 4

def has22(nums):
    for i in range(len(nums) - 1):
        if nums[i] == 2 and nums[i + 1] == 2:
            return True
    return False

This was the simplest solution I came up with.

Using a for loop to check if the iterated number, nums[i] == 2 "and" the one very next to it, [i+1] == 2 as well.

(len(nums)-1): this line prevents it from going out of the range through the for loop as the i+1 on the final loop will check out of the range.

Solution 5

def has22(nums):
    return '22' in ''.join(map(str, nums))

list [1, 2, 2] -> str '122'
return '22' in '122'

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