make_last

Given an int list, return a new list with double the length where its last element is the same as the original list, and all the other elements are 0. The original list will be length 1 or more. Note: by default, a new int list contains all 0’s.

make_last([4, 5, 6]) -> [0, 0, 0, 0, 0, 6]
make_last([1, 2]) -> [0, 0, 0, 2]
make_last([3]) -> [0, 3]

This exercise was taken from codingbat.com and has been adapted for the Python language. There are many great programming exercises there, but the majority are created for Java.

Starter Code

from typing import List


def make_last(nums: List[int]) -> List[int]:
    pass


result = make_last([4, 5, 6])
print(result)

Tests

from main import make_last


def test_make_last_1():
    assert make_last([4, 5, 6]) == [0, 0, 0, 0, 0, 6]


def test_make_last_2():
    assert make_last([1, 2]) == [0, 0, 0, 2]


def test_make_last_3():
    assert make_last([3]) == [0, 3]


def test_make_last_4():
    assert make_last([0]) == [0, 0]


def test_make_last_5():
    assert make_last([7, 7, 7]) == [0, 0, 0, 0, 0, 7]


def test_make_last_6():
    assert make_last([3, 1, 4]) == [0, 0, 0, 0, 0, 4]


def test_make_last_7():
    assert make_last([1, 2, 3, 4]) == [0, 0, 0, 0, 0, 0, 0, 4]


def test_make_last_8():
    assert make_last([1, 2, 3, 0]) == [0, 0, 0, 0, 0, 0, 0, 0]


def test_make_last_9():
    assert make_last([2, 4]) == [0, 0, 0, 4]