count_code

Requirements:

• substrings and slicing

• if, else

• loop through a string (while)

• loop with an accumulator variable

• returning a value

Return the number of times that the string “code” appears anywhere in the given string, except we’ll accept any letter for the ‘d’, so “cope” and “cooe” count.

count_code("aaacodebbb") -> 1
count_code("codexxcode") -> 2
count_code("cozexxcope") -> 2

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_code(string: str) -> int:
    pass


result = count_code('aaacodebbb')
print(result)

Tests

from main import count_code


def test_count_code_1():
    assert count_code('aaacodebbb') == 1


def test_count_code_2():
    assert count_code('codexxcode') == 2


def test_count_code_3():
    assert count_code('cozexxcope') == 2


def test_count_code_4():
    assert count_code('cozfxxcope') == 1


def test_count_code_5():
    assert count_code('xxcozeyycop') == 1


def test_count_code_6():
    assert count_code('cozcop') == 0


def test_count_code_7():
    assert count_code('abcxyz') == 0


def test_count_code_8():
    assert count_code('code') == 1


def test_count_code_9():
    assert count_code('ode') == 0


def test_count_code_10():
    assert count_code('c') == 0


def test_count_code_11():
    assert count_code('') == 0


def test_count_code_12():
    assert count_code('AAcodeBBcoleCCccoreDD') == 3


def test_count_code_13():
    assert count_code('AAcodeBBcoleCCccorfDD') == 2


def test_count_code_14():
    assert count_code('coAcodeBcoleccoreDD') == 3