in reply to Challenge: Sorting Sums Of Sorted Series
You can always save space by never storing anything  just recompute a sum every time you need it.. Here's a solution that uses constant space:
At each iteration, it traverses both lists to find the next smallest possible sum, and how many times it occurs. It only needs to keep track of $min, $nextmin, $multiplicity, $i, $j.sub print_sums { my ($listA, $listB) = @_; my $min = $listA>[0] + $listB>[0]  1; while ($min < $listA>[1] + $listB>[1]) { my $nextmin = undef; my $multiplicity = 0; for my $i (0 .. $#$listA) { for my $j (0 .. $#$listB) { my $sum = $listA>[$i] + $listB>[$j]; if ($sum > $min and ($sum < $nextmin or not defined $nextm +in)) { ($nextmin, $multiplicity) = ($sum, 1); } elsif ($sum == $nextmin) { $multiplicity++; } } } print( ($nextmin) x $multiplicity); $min = $nextmin; } }
Of course the tradeoff is the running time, which is O((NM)^{2}).
Perhaps the metric should be to minimize the product of time space complexities? For comparison, naively computing all sums and sorting uses NM space and NM log(NM) time, so it's slightly worse than mine under the combined metric.
Update: extra parens around print statement..
blokhead


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Re^2: Challenge: Sorting Sums Of Sorted Series
by Limbic~Region (Chancellor) on Feb 02, 2010 at 23:01 UTC  
by ikegami (Pope) on Feb 03, 2010 at 21:33 UTC  
by Limbic~Region (Chancellor) on Feb 03, 2010 at 21:50 UTC  
by ikegami (Pope) on Feb 03, 2010 at 22:18 UTC 
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