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| -rw-r--r-- | challenge-241/jeanluc2020/blog-1.txt | 1 | ||||
| -rw-r--r-- | challenge-241/jeanluc2020/blog-2.txt | 1 | ||||
| -rwxr-xr-x | challenge-241/jeanluc2020/perl/ch-1.pl | 69 | ||||
| -rwxr-xr-x | challenge-241/jeanluc2020/perl/ch-2.pl | 63 | ||||
| -rwxr-xr-x | challenge-241/jeanluc2020/python/ch-1.py | 60 | ||||
| -rwxr-xr-x | challenge-241/jeanluc2020/python/ch-2.py | 61 |
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diff --git a/challenge-241/jeanluc2020/blog-1.txt b/challenge-241/jeanluc2020/blog-1.txt new file mode 100644 index 0000000000..a86833251a --- /dev/null +++ b/challenge-241/jeanluc2020/blog-1.txt @@ -0,0 +1 @@ +http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-241-1.html diff --git a/challenge-241/jeanluc2020/blog-2.txt b/challenge-241/jeanluc2020/blog-2.txt new file mode 100644 index 0000000000..6ada722d0c --- /dev/null +++ b/challenge-241/jeanluc2020/blog-2.txt @@ -0,0 +1 @@ +http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-241-2.html diff --git a/challenge-241/jeanluc2020/perl/ch-1.pl b/challenge-241/jeanluc2020/perl/ch-1.pl new file mode 100755 index 0000000000..6ce8c5dffa --- /dev/null +++ b/challenge-241/jeanluc2020/perl/ch-1.pl @@ -0,0 +1,69 @@ +#!/usr/bin/perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-241/#TASK1 +# +# Task 1: Arithmetic Triplets +# =========================== +# +# You are given an array (3 or more members) of integers in increasing order +# and a positive integer. +# +# Write a script to find out the number of unique Arithmetic Triplets +# satisfying the following rules: +# +# a) i < j < k +# b) nums[j] - nums[i] == diff +# c) nums[k] - nums[j] == diff +# +## Example 1 +## +## Input: @nums = (0, 1, 4, 6, 7, 10) +## $diff = 3 +## Output: 2 +## +## Index (1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3. +## Index (2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3. +# +## Example 2 +## +## Input: @nums = (4, 5, 6, 7, 8, 9) +## $diff = 2 +## Output: 2 +## +## (0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2. +## (1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2. +# +############################################################ +## +## discussion +## +############################################################ +# +# Looping over the indices i, j, k we just check the condition. +# We can even short-curcuit the inner loop because we only need +# to execute it if $nums[$j] - $nums[$i] == $diff. + +use strict; +use warnings; + +arithmetic_triplets( [0, 1, 4, 6, 7, 10], 3 ); +arithmetic_triplets( [4, 5, 6, 7, 8, 9], 2); + +sub arithmetic_triplets { + my ($nums, $diff) = @_; + my @nums = @$nums; + my $result = 0; + print "Input: (" . join(", ", @nums) . "), $diff\n"; + foreach my $i (0..$#nums) { + foreach my $j ($i+1..$#nums) { + if($nums[$j] - $nums[$i] == $diff) { + foreach my $k ($j+1..$#nums) { + if($nums[$k] - $nums[$j] == $diff) { + $result++; + } + } + } + } + } + print "Output: $result\n"; +} + diff --git a/challenge-241/jeanluc2020/perl/ch-2.pl b/challenge-241/jeanluc2020/perl/ch-2.pl new file mode 100755 index 0000000000..f9a9dcbd36 --- /dev/null +++ b/challenge-241/jeanluc2020/perl/ch-2.pl @@ -0,0 +1,63 @@ +#!/usr/bin/perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-241/#TASK2 +# +# Task 2: Prime Order +# =================== +# +# You are given an array of unique positive integers greater than 2. +# +# Write a script to sort them in ascending order of the count of their prime +# factors, tie-breaking by ascending value. +# +## Example 1 +## +## Input: @int = (11, 8, 27, 4) +## Output: (11, 4, 8, 27)) +## +## Prime factors of 11 => 11 +## Prime factors of 4 => 2, 2 +## Prime factors of 8 => 2, 2, 2 +## Prime factors of 27 => 3, 3, 3 +# +############################################################ +## +## discussion +## +############################################################ +# +# Sort by number of prime factors first, then by the number +# itself second. +# We calculate the number of prime factors in an extra recursive +# function, using a cache of already calculated numbers to mitigate +# unnecessary recalculations +# +use strict; +use warnings; + +prime_order(11, 8, 27, 4); + +sub prime_order { + my @int = @_; + print "Input: (" . join(", ", @int) . ")\n"; + my @result = sort { prime_factor_count($a) <=> prime_factor_count($b) || $a <=> $b } @int; + print "Output: (" . join(", ", @result) . ")\n"; +} + +{ + my $cache = {}; + sub prime_factor_count { + my $number = shift; + my $count = 0; + unless($cache->{$number}) { + foreach my $i (2..$number) { + if($number % $i == 0) { + $count = 1 + prime_factor_count($number / $i); + last; + } + } + $cache->{$number} = $count; + } + + return $cache->{$number}; + } +} diff --git a/challenge-241/jeanluc2020/python/ch-1.py b/challenge-241/jeanluc2020/python/ch-1.py new file mode 100755 index 0000000000..4a96800f8d --- /dev/null +++ b/challenge-241/jeanluc2020/python/ch-1.py @@ -0,0 +1,60 @@ +#!/usr/bin/python3 +# https://theweeklychallenge.org/blog/perl-weekly-challenge-241/#TASK1 +# +# Task 1: Arithmetic Triplets +# =========================== +# +# You are given an array (3 or more members) of integers in increasing order +# and a positive integer. +# +# Write a script to find out the number of unique Arithmetic Triplets +# satisfying the following rules: +# +# a) i < j < k +# b) nums[j] - nums[i] == diff +# c) nums[k] - nums[j] == diff +# +## Example 1 +## +## Input: @nums = (0, 1, 4, 6, 7, 10) +## $diff = 3 +## Output: 2 +## +## Index (1, 2, 4) is an arithmetic triplet because both 7 - 4 == 3 and 4 - 1 == 3. +## Index (2, 4, 5) is an arithmetic triplet because both 10 - 7 == 3 and 7 - 4 == 3. +# +## Example 2 +## +## Input: @nums = (4, 5, 6, 7, 8, 9) +## $diff = 2 +## Output: 2 +## +## (0, 2, 4) is an arithmetic triplet because both 8 - 6 == 2 and 6 - 4 == 2. +## (1, 3, 5) is an arithmetic triplet because both 9 - 7 == 2 and 7 - 5 == 2. +# +############################################################ +## +## discussion +## +############################################################ +# +# Looping over the indices i, j, k we just check the condition. +# We can even short-curcuit the inner loop because we only need +# to execute it if $nums[$j] - $nums[$i] == $diff. + + +def arithmetic_triplets(nums: list, diff: int): + print("Input: (", ", ".join(str(x) for x in nums), "), ", diff, sep="") + result = 0 + for i in range(len(nums)): + for j in range(i+1, len(nums)): + if nums[j] - nums[i] == diff: + for k in range(j+1, len(nums)): + if nums[k] - nums[j] == diff: + result += 1 + print("Output:", result) + +arithmetic_triplets( [0, 1, 4, 6, 7, 10], 3 ); +arithmetic_triplets( [4, 5, 6, 7, 8, 9], 2); + + diff --git a/challenge-241/jeanluc2020/python/ch-2.py b/challenge-241/jeanluc2020/python/ch-2.py new file mode 100755 index 0000000000..507b88260b --- /dev/null +++ b/challenge-241/jeanluc2020/python/ch-2.py @@ -0,0 +1,61 @@ +#!/usr/bin/python3 +# https://theweeklychallenge.org/blog/perl-weekly-challenge-241/#TASK2 +# +# Task 2: Prime Order +# =================== +# +# You are given an array of unique positive integers greater than 2. +# +# Write a script to sort them in ascending order of the count of their prime +# factors, tie-breaking by ascending value. +# +## Example 1 +## +## Input: @int = (11, 8, 27, 4) +## Output: (11, 4, 8, 27)) +## +## Prime factors of 11 => 11 +## Prime factors of 4 => 2, 2 +## Prime factors of 8 => 2, 2, 2 +## Prime factors of 27 => 3, 3, 3 +# +############################################################ +## +## discussion +## +############################################################ +# +# Sort by number of prime factors first, then by the number +# itself second. +# We calculate the number of prime factors in an extra recursive +# function, using a cache of already calculated numbers to mitigate +# unnecessary recalculations +# + +from operator import itemgetter + +def prime_factor_count(number: int) -> int: + cache = {} + def prime_factor_count_interal(num: int) -> int: + count = 0 + if num not in cache.keys(): + for i in range(2, num+1): + if num % i == 0: + count = 1 + prime_factor_count_interal(int(num / i)) + break + cache[num] = count + return cache[num] + return prime_factor_count_interal(number) + + +def prime_order(nums: list): + print("Input: (", ", ".join(str(x) for x in nums), ")", sep='') + decorated = [ (prime_factor_count(x), x) for x in nums ] + decorated_result = sorted(decorated, key=lambda x: (x[0], x[1])) + result = [ x[1] for x in decorated_result ] + print("Output: (", ", ".join(str(x) for x in result), ")", sep='') + + +prime_order([11, 8, 27, 4]) +prime_order([11, 27, 8, 4]) + |