2015-07-04 07:57:09 +08:00
{
" cells " : [
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
2016-07-31 20:11:18 +08:00
" This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges). "
2015-07-04 07:57:09 +08:00
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" # Challenge Notebook "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Problem: Implement a function to reverse a string (a list of characters), in-place. \n " ,
" \n " ,
" * [Constraints](#Constraints) \n " ,
" * [Test Cases](#Test-Cases) \n " ,
" * [Algorithm](#Algorithm) \n " ,
" * [Code](#Code) \n " ,
" * [Unit Test](#Unit-Test) \n " ,
" * [Solution Notebook](#Solution-Notebook) "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Constraints \n " ,
" \n " ,
2016-06-13 11:06:12 +08:00
" * Can we assume the string is ASCII? \n " ,
2015-07-04 07:57:09 +08:00
" * Yes \n " ,
" * Note: Unicode strings could require special handling depending on your language \n " ,
" * Since we need to do this in-place, it seems we cannot use the slice operator or the reversed function? \n " ,
" * Correct \n " ,
2016-06-13 11:06:12 +08:00
" * Since Python string are immutable, can we use a list of characters instead? \n " ,
2015-07-04 07:57:09 +08:00
" * Yes "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Test Cases \n " ,
" \n " ,
" * None -> None \n " ,
" * [ ' ' ] -> [ ' ' ] \n " ,
" * [ ' f ' , ' o ' , ' o ' , ' ' , ' b ' , ' a ' , ' r ' ] -> [ ' r ' , ' a ' , ' b ' , ' ' , ' o ' , ' o ' , ' f ' ] "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Algorithm \n " ,
" \n " ,
2015-07-05 19:59:00 +08:00
" Refer to the [Solution Notebook](http://nbviewer.ipython.org/github/donnemartin/interactive-coding-challenges/blob/master/arrays_strings/reverse_string/reverse_string_solution.ipynb). If you are stuck and need a hint, the solution notebook ' s algorithm discussion might be a good place to start. "
2015-07-04 07:57:09 +08:00
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Code "
]
} ,
{
" cell_type " : " code " ,
" execution_count " : null ,
" metadata " : {
" collapsed " : false
} ,
" outputs " : [ ] ,
" source " : [
2016-02-14 19:04:48 +08:00
" def reverse_string(list_chars): \n " ,
2015-07-04 07:57:09 +08:00
" # TODO: Implement me \n " ,
" pass "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Unit Test "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" \n " ,
" \n " ,
" **The following unit test is expected to fail until you solve the challenge.** "
]
} ,
{
" cell_type " : " code " ,
" execution_count " : null ,
" metadata " : {
" collapsed " : false
} ,
" outputs " : [ ] ,
" source " : [
" # %lo ad test_reverse_string.py \n " ,
" from nose.tools import assert_equal \n " ,
" \n " ,
" \n " ,
" class TestReverse(object): \n " ,
2015-07-12 03:34:14 +08:00
" \n " ,
2015-07-04 07:57:09 +08:00
" def test_reverse(self): \n " ,
2016-02-14 19:04:48 +08:00
" assert_equal(reverse_string(None), None) \n " ,
" assert_equal(reverse_string([ ' ' ]), [ ' ' ]) \n " ,
" assert_equal(reverse_string( \n " ,
2015-07-12 03:34:14 +08:00
" [ ' f ' , ' o ' , ' o ' , ' ' , ' b ' , ' a ' , ' r ' ]), \n " ,
" [ ' r ' , ' a ' , ' b ' , ' ' , ' o ' , ' o ' , ' f ' ]) \n " ,
2015-07-04 07:57:09 +08:00
" print( ' Success: test_reverse ' ) \n " ,
" \n " ,
2016-01-21 02:40:10 +08:00
" def test_reverse_inplace(self): \n " ,
" target_list = [ ' f ' , ' o ' , ' o ' , ' ' , ' b ' , ' a ' , ' r ' ] \n " ,
2016-02-14 19:04:48 +08:00
" reverse_string(target_list) \n " ,
2016-01-21 02:40:10 +08:00
" assert_equal(target_list, [ ' r ' , ' a ' , ' b ' , ' ' , ' o ' , ' o ' , ' f ' ]) \n " ,
" print( ' Success: test_reverse_inplace ' ) \n " ,
" \n " ,
2015-07-12 03:34:14 +08:00
" \n " ,
2015-07-04 07:57:09 +08:00
" def main(): \n " ,
" test = TestReverse() \n " ,
" test.test_reverse() \n " ,
2016-01-21 02:40:10 +08:00
" test.test_reverse_inplace() \n " ,
2015-07-12 03:34:14 +08:00
" \n " ,
" \n " ,
2015-07-04 07:57:09 +08:00
" if __name__ == ' __main__ ' : \n " ,
" main() "
]
} ,
{
" cell_type " : " markdown " ,
" metadata " : { } ,
" source " : [
" ## Solution Notebook \n " ,
" \n " ,
2015-07-05 19:59:00 +08:00
" Review the [Solution Notebook](http://nbviewer.ipython.org/github/donnemartin/interactive-coding-challenges/blob/master/arrays_strings/reverse_string/reverse_string_solution.ipynb) for a discussion on algorithms and code solutions. "
2015-07-04 07:57:09 +08:00
]
}
] ,
" metadata " : {
" kernelspec " : {
2016-02-14 19:04:48 +08:00
" display_name " : " Python 3 " ,
2015-07-04 07:57:09 +08:00
" language " : " python " ,
2016-02-14 19:04:48 +08:00
" name " : " python3 "
2015-07-04 07:57:09 +08:00
} ,
" language_info " : {
" codemirror_mode " : {
" name " : " ipython " ,
2016-02-14 19:04:48 +08:00
" version " : 3
2015-07-04 07:57:09 +08:00
} ,
" file_extension " : " .py " ,
" mimetype " : " text/x-python " ,
" name " : " python " ,
" nbconvert_exporter " : " python " ,
2016-02-14 19:04:48 +08:00
" pygments_lexer " : " ipython3 " ,
2016-06-13 11:06:12 +08:00
" version " : " 3.5.0 "
2015-07-04 07:57:09 +08:00
}
} ,
" nbformat " : 4 ,
" nbformat_minor " : 0
}