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  1. interactive-coding-challenges
  2. arrays_strings
  3. rotation

This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Challenge Notebook¶

Problem: Determine if a string s1 is a rotation of another string s2, by calling (only once) a function is_substring.¶

  • Constraints
  • Test Cases
  • Algorithm
  • Code
  • Unit Test
  • Solution Notebook

Constraints¶

  • Can we assume the string is ASCII?
    • Yes
    • Note: Unicode strings could require special handling depending on your language
  • Is this case sensitive?
    • Yes
  • Can we use additional data structures?
    • Yes
  • Can we assume this fits in memory?
    • Yes

Test Cases¶

  • Any strings that differ in size -> False
  • None, 'foo' -> False (any None results in False)
  • ' ', 'foo' -> False
  • ' ', ' ' -> True
  • 'foobarbaz', 'barbazfoo' -> True

Algorithm¶

Refer to the Solution Notebook. If you are stuck and need a hint, the solution notebook's algorithm discussion might be a good place to start.

Code¶

In [ ]:
class Rotation(object):

    def is_substring(self, s1, s2):
        # TODO: Implement me
        pass

    def is_rotation(self, s1, s2):
        # TODO: Implement me
        # Call is_substring only once
        pass

Unit Test¶

The following unit test is expected to fail until you solve the challenge.

In [ ]:
# %load test_rotation.py
import unittest


class TestRotation(unittest.TestCase):

    def test_rotation(self):
        rotation = Rotation()
        self.assertEqual(rotation.is_rotation('o', 'oo'), False)
        self.assertEqual(rotation.is_rotation(None, 'foo'), False)
        self.assertEqual(rotation.is_rotation('', 'foo'), False)
        self.assertEqual(rotation.is_rotation('', ''), True)
        self.assertEqual(rotation.is_rotation('foobarbaz', 'barbazfoo'), True)
        print('Success: test_rotation')


def main():
    test = TestRotation()
    test.test_rotation()


if __name__ == '__main__':
    main()

Solution Notebook¶

Review the Solution Notebook for a discussion on algorithms and code solutions.

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