#! /bin/perl -CSDA =pod ------------------------------------------------------------------------------------------------------------------------ COLOPHON: This is a 120-character-wide Unicode UTF-8 Perl-source-code text file with hard Unix line breaks ("\x0A"). ¡Hablo Español! Говорю Русский. Björt skjöldur. ॐ नमो भगवते वासुदेवाय. 看的星星，知道你是爱。麦藁雪、富士川町、山梨県。 ------------------------------------------------------------------------------------------------------------------------ TITLE BLOCK: ch-2.pl Robbie Hatley's Perl solutions for The Weekly Challenge 215-2. Written by Robbie Hatley on Tue May 2, 2023. ------------------------------------------------------------------------------------------------------------------------ Task 2: Number Placement Submitted by: Mohammad S Anwar You are given a list of numbers having just 0 and 1. You are also given placement count (>=1). Write a script to find out if it is possible to replace 0 with 1 the given number of times in the given list. The only condition is that you can only replace when there is no 1 on either side. Print 1 if it is possible otherwise 0. Example 1: Input: @numbers = (1,0,0,0,1), $count = 1 Output: 1 Example 2: Input: @numbers = (1,0,0,0,1), $count = 2 Output: 0 Example 3: Input: @numbers = (1,0,0,0,0,0,0,0,1), $count = 3 Output: 1 ------------------------------------------------------------------------------------------------------------------------ PROBLEM NOTES: Conceptually, this is even simpler than Task 1: Just replace what we can, keep tally, and if tally >= $count, print 1, else 0. But in-practice, there is the annoying matter of "special cases": Arrays of sizes 0,1,2 and the 0th and last elements of every array. Those all need to be handled separately, so it's not just a case of writing a sub and applying it to every element of an array using a for loop. In fact, I ended up using NO subs but a disgusting number of "if" statements to handle the special cases. ------------------------------------------------------------------------------------------------------------------------ INPUT / OUTPUT NOTES: Input is via either built-in variables or via @ARGV. If using @ARGV, provide one argument which must be a single-quoted array of arrays in proper Perl syntax, with each inner array being a list of 1s and 0s followed by count. Eg: ./ch-2.pl '([1,0,1,0,1,1],[1,0,0,0,2],[1,0,0,0,1],[0,1,0,0,0,0,0,0,0,0,1,2])' Output is to STDOUT and will be list of numbers, count, and number of replacements, on separate lines. =cut # ---------------------------------------------------------------------------------------------------------------------- # PRELIMINARIES: use v5.36; use strict; use warnings; use utf8; use Sys::Binmode; use Time::HiRes 'time'; $"=', '; # ---------------------------------------------------------------------------------------------------------------------- # DEFAULT INPUTS: my @arrays = ( [1,0,0,0,1,1], [1,0,0,0,1,2], [1,0,0,0,0,0,0,0,1,3], ); # ---------------------------------------------------------------------------------------------------------------------- # NON-DEFAULT INPUTS: if (@ARGV) {@arrays = eval($ARGV[0]);} # ---------------------------------------------------------------------------------------------------------------------- # MAIN BODY OF PROGRAM: { # begin main my $t0 = time; for my $aref (@arrays) { my @numbers = @$aref; # "Original" copy of array. my $count = pop @numbers; # Pop "desired replacement count" off end of @numbers. my @mutable = @numbers; # "Working" copy of @numbers; this may get altered. my $yes = 0; # WAS @mutable altered? my $tally = 0; # If so, how many TIMES was it altered? my $sz = scalar(@mutable); # Size of @mutable. # First of all, we need to handle three bloody-annoying "special-case" array sizes: 0, 1, 2: if ( 0 == $sz ) { # If array has 0 elements: ; # No replacements are possible, so do nothing. } elsif ( 1 == $sz ) { # Else if array has 1 element: if ( $mutable[0] == 0 ) { # If element is 0, $mutable[0] = 1; # replace 0 with 1 ++$tally; # and increment tally. } } elsif ( 2 == $sz ) { # Else if array has 2 elements: if ( $mutable[0] == 0 # If 0th element is 0 && $mutable[1] != 1 ) { # and 1st element is not 1, $mutable[0] = 1; # replace 0 with 1 ++$tally; # and increment tally. } if ( $mutable[1] == 0 # If 1st element is 0, && $mutable[0] != 1 ) { # and 0th element is not 1 $mutable[1] = 1; # replace 0 with 1 ++$tally; # and increment tally. } } # at-most-one 0 can be replaced by 1.) # Now, finally, we can get on with the code for the vast majority of cases, in which $sz > 2: else { # Else array has 3-or-more elements: if ( $mutable[0] == 0 # If 0th element is 0, && $mutable[1] != 1) { # and 1st element is not 1, $mutable[0] = 1; # replace 0 with 1 ++$tally; # and increment tally. } for ( my $i = 1 ; $i <= $#mutable-1 ; ++$i ) { # For each non-end element, if ( $mutable[$i] == 0 # if it's 0 && $mutable[$i-1] != 1 # and element to left is not 1 && $mutable[$i+1] != 1 ) { # and element to right is not 1 $mutable[$i] = 1; # replace 0 with 1 ++$tally; # and increment tally. } } if ( $mutable[$#mutable] == 0 # If last element is 0, && $mutable[$#mutable-1] != 1 ) { # and penultimate element is not 1, $mutable[$#mutable] = 1; # replace 0 with 1 ++$tally; # and increment tally. } } if ($tally >= $count) { # If we were able to make AT-LEAST as many replacements $yes = 1; # as were requested, then set $yes to 1; } # else leave it set to 0. say ''; # Print results: say "numbers: (@numbers)"; # Original input array. say "quota: $count"; # Requested replacement count. say "replaced: $tally"; # Tally of actual replacements. say "met quota?: $yes"; # Output. } my $µs = 1000000 * (time - $t0); # Calculate elpased time in µs, printf("\nExecution time was %.3fµs.\n", $µs); # print elapsed time, exit 0; # and exit program. } # end main