diff options
Diffstat (limited to 'challenge-063')
| -rw-r--r-- | challenge-063/paulo-custodio/Makefile | 1 | ||||
| -rw-r--r-- | challenge-063/paulo-custodio/perl/ch-1.pl | 4 | ||||
| -rw-r--r-- | challenge-063/paulo-custodio/perl/ch-2.pl | 5 | ||||
| -rw-r--r-- | challenge-063/paulo-custodio/python/ch-1.py | 45 | ||||
| -rw-r--r-- | challenge-063/paulo-custodio/python/ch-2.py | 47 |
5 files changed, 98 insertions, 4 deletions
diff --git a/challenge-063/paulo-custodio/Makefile b/challenge-063/paulo-custodio/Makefile index 0d5753b0fb..3eba7cf291 100644 --- a/challenge-063/paulo-custodio/Makefile +++ b/challenge-063/paulo-custodio/Makefile @@ -1,3 +1,4 @@ all: perl perl/ch-1.pl + python3 python/ch-1.py perl ../../challenge-001/paulo-custodio/test.pl diff --git a/challenge-063/paulo-custodio/perl/ch-1.pl b/challenge-063/paulo-custodio/perl/ch-1.pl index d3900d26b7..bf8874c318 100644 --- a/challenge-063/paulo-custodio/perl/ch-1.pl +++ b/challenge-063/paulo-custodio/perl/ch-1.pl @@ -2,7 +2,7 @@ # Challenge 063 # -# TASK #1 › Last Word +# TASK #1 > Last Word # Submitted by: Mohammad S Anwar # Reviewed by: Ryan Thompson # @@ -10,7 +10,7 @@ # $regexp found in the given string, or undef if the string does not contain a # word matching $regexp. # -# For this challenge, a “word” is defined as any character sequence consisting +# For this challenge, a "word" is defined as any character sequence consisting # of non-whitespace characters (\S) only. That means punctuation and other # symbols are part of the word. # diff --git a/challenge-063/paulo-custodio/perl/ch-2.pl b/challenge-063/paulo-custodio/perl/ch-2.pl index 70ad2a5098..c477a46a40 100644 --- a/challenge-063/paulo-custodio/perl/ch-2.pl +++ b/challenge-063/paulo-custodio/perl/ch-2.pl @@ -2,11 +2,12 @@ # Challenge 063 # -# TASK #2 › Rotate String +# TASK #2 > Rotate String # Submitted by: Mohammad S Anwar # Reviewed by: Ryan Thompson # -# Given a word made up of an arbitrary number of x and y characters, that word can be rotated as follows: For the ith rotation (starting at i = 1), +# Given a word made up of an arbitrary number of x and y characters, that word +# can be rotated as follows: For the ith rotation (starting at i = 1), # i % length(word) characters are moved from the front of the string to the end. # Thus, for the string xyxx, the initial (i = 1) % 4 = 1 character (x) is moved # to the end, forming yxxx. On the second rotation, (i = 2) % 4 = 2 characters diff --git a/challenge-063/paulo-custodio/python/ch-1.py b/challenge-063/paulo-custodio/python/ch-1.py new file mode 100644 index 0000000000..5fb968ca0d --- /dev/null +++ b/challenge-063/paulo-custodio/python/ch-1.py @@ -0,0 +1,45 @@ +#!/usr/bin/env python3 + +# Challenge 063 +# +# TASK #1 > Last Word +# Submitted by: Mohammad S Anwar +# Reviewed by: Ryan Thompson +# +# Define sub last_word($string, $regexp) that returns the last word matching +# $regexp found in the given string, or undef if the string does not contain a +# word matching $regexp. +# +# For this challenge, a "word" is defined as any character sequence consisting +# of non-whitespace characters (\S) only. That means punctuation and other +# symbols are part of the word. +# +# The $regexp is a regular expression. Take care that the regexp can only match +# individual words! See the Examples for one way this can break if you are not +# careful. +# +# Examples +# last_word(' hello world', qr/[ea]l/); # 'hello' +# last_word("Don't match too much, Chet!", qr/ch.t/i); # 'Chet!' +# last_word("spaces in regexp won't match", qr/in re/); # undef +# last_word( join(' ', 1..1e6), qr/^(3.*?){3}/); # '399933' + +import re +import unittest + +def last_word(text, pattern): + words = text.split(' ') + for word in reversed(words): + if re.search(pattern, word, re.IGNORECASE): + return word + return None + +class TestLastWord(unittest.TestCase): + def test_last_word(self): + self.assertEqual(last_word(' hello world', r'[ea]l'), 'hello') + self.assertEqual(last_word("Don't match too much, Chet!", r'ch.t'), 'Chet!') + self.assertIsNone(last_word("spaces in regexp won't match", r'in re')) + self.assertEqual(last_word(' '.join(str(i) for i in range(1, int(1e6) + 1)), r'^(3.*?){3}'), '399933') + +if __name__ == '__main__': + unittest.main() diff --git a/challenge-063/paulo-custodio/python/ch-2.py b/challenge-063/paulo-custodio/python/ch-2.py new file mode 100644 index 0000000000..37f0bab8d2 --- /dev/null +++ b/challenge-063/paulo-custodio/python/ch-2.py @@ -0,0 +1,47 @@ +#!/usr/bin/env python3 + +# Challenge 063 +# +# TASK #2 > Rotate String +# Submitted by: Mohammad S Anwar +# Reviewed by: Ryan Thompson +# +# Given a word made up of an arbitrary number of x and y characters, that word +# can be rotated as follows: For the ith rotation (starting at i = 1), +# i % length(word) characters are moved from the front of the string to the end. +# Thus, for the string xyxx, the initial (i = 1) % 4 = 1 character (x) is moved +# to the end, forming yxxx. On the second rotation, (i = 2) % 4 = 2 characters +# (yx) are moved to the end, forming xxyx, and so on. See below for a complete +# example. +# +# Your task is to write a function that takes a string of xs and ys and returns +# the minimum non-zero number of rotations required to obtain the original +# string. You may show the individual rotations if you wish, but that is not +# required. +# +# Example +# Input: $word = 'xyxx'; +# +# Rotation 1: you get yxxx by moving x to the end. +# Rotation 2: you get xxyx by moving yx to the end. +# Rotation 3: you get xxxy by moving xxy to the end. +# Rotation 4: you get xxxy by moving nothing as 4 % length(xyxx) == 0. +# Rotation 5: you get xxyx by moving x to the end. +# Rotation 6: you get yxxx by moving xx to the end. +# Rotation 7: you get xyxx by moving yxx to the end which is same as the given word. +# Output: 7 + +import sys + +def count_rotations(in_str=""): + count = 0 + str_val = in_str + while True: + count += 1 + move = count % len(str_val) + str_val = str_val[move:] + str_val[:move] + if in_str == str_val: + break + return count + +print(count_rotations(sys.argv[1] if len(sys.argv) > 1 else "")) |