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| -rw-r--r-- | challenge-248/jeanluc2020/blog-1.txt | 1 | ||||
| -rw-r--r-- | challenge-248/jeanluc2020/blog-2.txt | 1 | ||||
| -rwxr-xr-x | challenge-248/jeanluc2020/perl/ch-1.pl | 70 | ||||
| -rwxr-xr-x | challenge-248/jeanluc2020/perl/ch-2.pl | 74 | ||||
| -rwxr-xr-x | challenge-248/jeanluc2020/python/ch-1.py | 66 | ||||
| -rwxr-xr-x | challenge-248/jeanluc2020/python/ch-2.py | 71 |
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diff --git a/challenge-248/jeanluc2020/blog-1.txt b/challenge-248/jeanluc2020/blog-1.txt new file mode 100644 index 0000000000..6f63c724bf --- /dev/null +++ b/challenge-248/jeanluc2020/blog-1.txt @@ -0,0 +1 @@ +http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-248-1.html diff --git a/challenge-248/jeanluc2020/blog-2.txt b/challenge-248/jeanluc2020/blog-2.txt new file mode 100644 index 0000000000..d66dd45736 --- /dev/null +++ b/challenge-248/jeanluc2020/blog-2.txt @@ -0,0 +1 @@ +http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-248-2.html diff --git a/challenge-248/jeanluc2020/perl/ch-1.pl b/challenge-248/jeanluc2020/perl/ch-1.pl new file mode 100755 index 0000000000..d843d3e10c --- /dev/null +++ b/challenge-248/jeanluc2020/perl/ch-1.pl @@ -0,0 +1,70 @@ +#!/usr/bin/env perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-248/#TASK1 +# +# Task 1: Shortest Distance +# ========================= +# +# You are given a string and a character in the given string. +# +# Write a script to return an array of integers of size same as length of the +# given string such that: +# +# distance[i] is the distance from index i to the closest occurence of +# the given character in the given string. +# +# The distance between two indices i and j is abs(i - j). +# +## Example 1 +## +## Input: $str = "loveleetcode", $char = "e" +## Output: (3,2,1,0,1,0,0,1,2,2,1,0) +## +## The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed). +## The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3. +## The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2. +## For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, +## but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1. +## The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2. +# +## Example 2 +## +## Input: $str = "aaab", $char = "b" +## Output: (3,2,1,0) +# +############################################################ +## +## discussion +## +############################################################ +# +# Let's split the input string into its characters and fill an +# array from that. Then we create a map using the positions of +# all occurrences of the wanted character. Now we walk the array +# and take the minimum of all possible differences found to +# elements from the map, or -1 if the map is empty (this is +# just extra error handling in case someone throws a character +# into the function that doesn't exist in the string, we just +# return -1 for all characters in that case). + +use strict; +use warnings; +use List::Util qw(min); + +shortest_distance("loveleetcode", "e"); +shortest_distance("aaab", "b"); +shortest_distance("aaab", "c"); + +sub shortest_distance { + my ($string, $char) = @_; + print "Input: '$string', '$char'\n"; + my @str_chars = split //, $string; + my @map; + foreach my $i (0..$#str_chars) { + push @map, $i if $str_chars[$i] eq $char; + } + my @result = (); + foreach my $i (0..$#str_chars) { + push @result, min( map { abs($i - $_), } @map ) // -1; + } + print "Output: (" . join(",", @result) . ")\n"; +} diff --git a/challenge-248/jeanluc2020/perl/ch-2.pl b/challenge-248/jeanluc2020/perl/ch-2.pl new file mode 100755 index 0000000000..31fda73f4f --- /dev/null +++ b/challenge-248/jeanluc2020/perl/ch-2.pl @@ -0,0 +1,74 @@ +#!/usr/bin/env perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-248/#TASK2 +# +# Task 2: Submatrix Sum +# ===================== +# +# You are given a NxM matrix A of integers. +# +# Write a script to construct a (N-1)x(M-1) matrix B having elements that are +# the sum over the 2x2 submatrices of A, +# +# b[i,k] = a[i,k] + a[i,k+1] + a[i+1,k] + a[i+1,k+1] +# +# +## Example 1 +## +## Input: $a = [ +## [1, 2, 3, 4], +## [5, 6, 7, 8], +## [9, 10, 11, 12] +## ] +## +## Output: $b = [ +## [14, 18, 22], +## [30, 34, 38] +## ] +# +## Example 2 +## +## Input: $a = [ +## [1, 0, 0, 0], +## [0, 1, 0, 0], +## [0, 0, 1, 0], +## [0, 0, 0, 1] +## ] +## +## Output: $b = [ +## [2, 1, 0], +## [1, 2, 1], +## [0, 1, 2] +## ] +# +############################################################ +## +## discussion +## +############################################################ +# +# Just some index arithmetic + +submatrix_sum( [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12] ] ); +submatrix_sum( [ [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1] ] ); + +sub submatrix_sum { + my $matrix = shift; + print "Input: [\n"; + foreach my $array (@$matrix) { + print " [" . join(",", @$array) . "],\n"; + } + print " ]\n"; + my $l = scalar(@$matrix); + my $m = scalar(@{$matrix->[0]}); + my $resultmatrix = []; + foreach my $i (0..$l-2) { + foreach my $k (0..$m-2) { + $resultmatrix->[$i]->[$k] = $matrix->[$i]->[$k] + $matrix->[$i]->[$k+1] + $matrix->[$i+1]->[$k] + $matrix->[$i+1]->[$k+1]; + } + } + print "Output: [\n"; + foreach my $array (@$resultmatrix) { + print " [" . join(",", @$array) . "],\n"; + } + print " ]\n"; +} diff --git a/challenge-248/jeanluc2020/python/ch-1.py b/challenge-248/jeanluc2020/python/ch-1.py new file mode 100755 index 0000000000..30c51061e9 --- /dev/null +++ b/challenge-248/jeanluc2020/python/ch-1.py @@ -0,0 +1,66 @@ +#!/usr/bin/python3 +# https://theweeklychallenge.org/blog/perl-weekly-challenge-248/#TASK1 +# +# Task 1: Shortest Distance +# ========================= +# +# You are given a string and a character in the given string. +# +# Write a script to return an array of integers of size same as length of the +# given string such that: +# +# distance[i] is the distance from index i to the closest occurence of +# the given character in the given string. +# +# The distance between two indices i and j is abs(i - j). +# +## Example 1 +## +## Input: $str = "loveleetcode", $char = "e" +## Output: (3,2,1,0,1,0,0,1,2,2,1,0) +## +## The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed). +## The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3. +## The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2. +## For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, +## but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1. +## The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2. +# +## Example 2 +## +## Input: $str = "aaab", $char = "b" +## Output: (3,2,1,0) +# +############################################################ +## +## discussion +## +############################################################ +# +# Let's split the input string into its characters and fill an +# array from that. Then we create a map using the positions of +# all occurrences of the wanted character. Now we walk the array +# and take the minimum of all possible differences found to +# elements from the map, or -1 if the map is empty (this is +# just extra error handling in case someone throws a character +# into the function that doesn't exist in the string, we just +# return -1 for all characters in that case). + +def shortest_distance(string: str, char: str) -> list: + print(f"Input: '{string}', '{char}'") + str_chars = list(string) + map = [] + for i in range(len(str_chars)): + if str_chars[i] == char: + map.append(i) + result = [] + for i in range(len(str_chars)): + result.append(min([abs(i-x) for x in map], default = -1)) + print(f"Output: (", ",".join(str(x) for x in result), ")") + return result + + +shortest_distance("loveleetcode", "e") +shortest_distance("aaab", "b") +shortest_distance("aaab", "c") + diff --git a/challenge-248/jeanluc2020/python/ch-2.py b/challenge-248/jeanluc2020/python/ch-2.py new file mode 100755 index 0000000000..e2177fe4b0 --- /dev/null +++ b/challenge-248/jeanluc2020/python/ch-2.py @@ -0,0 +1,71 @@ +#!/usr/bin/env perl +# https://theweeklychallenge.org/blog/perl-weekly-challenge-248/#TASK2 +# +# Task 2: Submatrix Sum +# ===================== +# +# You are given a NxM matrix A of integers. +# +# Write a script to construct a (N-1)x(M-1) matrix B having elements that are +# the sum over the 2x2 submatrices of A, +# +# b[i,k] = a[i,k] + a[i,k+1] + a[i+1,k] + a[i+1,k+1] +# +# +## Example 1 +## +## Input: $a = [ +## [1, 2, 3, 4], +## [5, 6, 7, 8], +## [9, 10, 11, 12] +## ] +## +## Output: $b = [ +## [14, 18, 22], +## [30, 34, 38] +## ] +# +## Example 2 +## +## Input: $a = [ +## [1, 0, 0, 0], +## [0, 1, 0, 0], +## [0, 0, 1, 0], +## [0, 0, 0, 1] +## ] +## +## Output: $b = [ +## [2, 1, 0], +## [1, 2, 1], +## [0, 1, 2] +## ] +# +############################################################ +## +## discussion +## +############################################################ +# +# Just some index arithmetic + +def submatrix_sum(matrix: list) -> list: + print("Input: [") + for array in matrix: + print(" [", ",".join([str(x) for x in array]), "],") + print(" ]") + l = len(matrix) + m = len(matrix[0]) + resultmatrix = [] + for i in range(l-1): + resultmatrix.append([]) + for k in range(m-1): + resultmatrix[i].append( matrix[i][k] + matrix[i][k+1] + matrix[i+1][k] + matrix[i+1][k+1] ) + print("Output: [") + for array in resultmatrix: + print(" [", ",".join([str(x) for x in array]), "],") + print(" ]") + return resultmatrix + +submatrix_sum( [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12] ] ) +submatrix_sum( [ [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1] ] ) + |