zig_zag.py

"""
Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

v1 = [1, 2]
v2 = [3, 4, 5, 6]
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].

Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

Clarification for the follow up question - Update (2015-09-18):
The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:

[1,2,3]
[4,5,6,7]
[8,9]
It should return [1,4,8,2,5,9,3,6,7].
"""


from itertools import zip_longest

given = [[1, 2, 3], [4, 5, 6, 7], [8, 9]]
expected = [1, 4, 8, 2, 5, 9, 3, 6, 7]


def zig_zag(iters: [list]) -> list:
    return [x for x in sum(map(list, zip_longest(*iters)), []) if x is not None]


print(zig_zag(given))
assert zig_zag(given) == expected