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.

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.

```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];
}