prefix_again

Requirements:

• mathematical operations

• substrings and slicing

• loop with a counter variable

• if, else

• break

• returning a value

Given a string, consider the prefix string made of the first n chars of the string. Does that prefix string appear somewhere else in the string? Assume that the string is not empty and that n is in the range 1..len(str).

prefix_again("abXYabc", 1) -> true
prefix_again("abXYabc", 2) -> true
prefix_again("abXYabc", 3) -> false

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 prefix_again(string: str, n: int) -> bool:
    pass


result = prefix_again('abXYabc', 1)
print(result)

Tests

from main import prefix_again


def test_prefix_again_1():
    assert prefix_again('abXYabc', 1) == True


def test_prefix_again_2():
    assert prefix_again('abXYabc', 2) == True


def test_prefix_again_3():
    assert prefix_again('abXYabc', 3) == False


def test_prefix_again_4():
    assert prefix_again('xyzxyxyxy', 2) == True


def test_prefix_again_5():
    assert prefix_again('xyzxyxyxy', 3) == False


def test_prefix_again_6():
    assert prefix_again('Hi12345Hi6789Hi10', 1) == True


def test_prefix_again_7():
    assert prefix_again('Hi12345Hi6789Hi10', 2) == True


def test_prefix_again_8():
    assert prefix_again('Hi12345Hi6789Hi10', 3) == True


def test_prefix_again_9():
    assert prefix_again('Hi12345Hi6789Hi10', 4) == False


def test_prefix_again_10():
    assert prefix_again('a', 1) == False


def test_prefix_again_11():
    assert prefix_again('aa', 1) == True


def test_prefix_again_12():
    assert prefix_again('ab', 1) == False