count_8

Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while integer divide (//) by 10 removes the rightmost digit (126 // 10 is 12).

count_8(8) -> 1
count_8(818) -> 2
count_8(8818) -> 4

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

def count_8(n: int) -> int:
    pass


result = count_8(8)
print(result)

Tests

from main import count_8


def test_count_8_1():
    assert count_8(8) == 1


def test_count_8_2():
    assert count_8(818) == 2


def test_count_8_3():
    assert count_8(8818) == 4


def test_count_8_4():
    assert count_8(8088) == 4


def test_count_8_5():
    assert count_8(123) == 0


def test_count_8_6():
    assert count_8(81238) == 2


def test_count_8_7():
    assert count_8(88788) == 6


def test_count_8_8():
    assert count_8(8234) == 1


def test_count_8_9():
    assert count_8(2348) == 1


def test_count_8_10():
    assert count_8(23884) == 3


def test_count_8_11():
    assert count_8(0) == 0


def test_count_8_12():
    assert count_8(1818188) == 5


def test_count_8_13():
    assert count_8(8818181) == 5


def test_count_8_14():
    assert count_8(1080) == 1


def test_count_8_15():
    assert count_8(188) == 3


def test_count_8_16():
    assert count_8(88888) == 9


def test_count_8_17():
    assert count_8(9898) == 2


def test_count_8_18():
    assert count_8(78) == 1