in reply to Re: partition of an array
in thread partition of an array
For the universe of positive integers, I sensed a simple solution lurking. I found one but I didn't like how it
coded up: It was just pushing the largest onto a half,
if it was forced onto the currently smaller half, put it
in a protest queue in that half. When ready to add a number to the other half in the presence of a protest, you would instead pop the top of the queue to the other half. This works and is fairly efficient but it is messy in the control code.
rir
Today, it just struck me that you don't have to keep the halves abs( @left - @right) <= 1 as you create them; just make sure there is enough remaining to fill the other half.
Be well,use List::Util qw{ sum }; sub L() { 0} # left sub R() { 1} # right sub remainder_halves { my $in = shift; my $ar ; @$ar = sort { $b <=> $a } @$in; die "bounds error" if @$ar && $$ar[-1] < 0; my @ans = ( [], [] ); # halves for answer no warnings 'uninitialized'; # summing empty arrays my ( $targ, $other ) = ( L, R ); my ( $halfsize) = int((@$ar+1)/2); while ( @$ar ) { while ( sum( @{$ans[$targ]}) <= sum( @{$ans[$other]}) && @{$ans[$targ]} < $halfsize ) { push @{$ans[$targ]}, shift @$ar; } ( $targ, $other ) = ( $other, $targ); push @{$ans[$targ]}, shift @$ar if @{$ans[$targ]} < $halfsize; } my $score = abs( sum( @{ $ans[L] } ) - sum( @{ $ans[R] } ) ); return $score, $ans[L], $ans[R]; }
rir
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