Golfed, I get you down to 140 , assuming the inputs are ordered smallest to largest. Interestingly enough, if you have the defined-or patch, I get down to 122. If the inputs aren't ordered, add 13 characters to both solutions.
Without //= patch:
@x=@_;my%c;$e=sub{my$v=pop;exists$c{$v}?$c{$v}:$c{$v}=$v<0?0:$v==0||gr
+ep&$e($v-$_),@x};$t=$s=0;{&$e(++$t)?$t-$s>=$x[0]&&last:($s=$t);redo}$
+s
With //= patch:
@x=@_;my%c;$e=sub{my$v=pop;$c{$v}//=$v<0?0:$v==0||grep&$e($v-$_),@x};$
+t=$s=0;{&$e(++$t)?$t-$s>=$x[0]&&last:($s=$t);redo}$s
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Update: Actually, the inputs don't have to be ordered. It just means that the algorithm will take a little longer, but it will get the right results. Also, drop a character by reordering the assignment to $c{$v}. The //= patched version looks like:
@x=@_;my%c;$e=sub{my$v=pop;$c{$v}//=$v==0||$v>0&&grep&$e($v-$_),@x};$t
+=$s=0;{&$e(++$t)?$t-$s>=$x[0]&&last:($s=$t);redo}$s
|
Update: Rewriting the redo-loop as a C-style for-loop drops to 111 characters for the //= patched version.
@x=@_;my%c;$e=sub{my$v=pop;$c{$v}//=!$v||$v>0&&grep&$e($v-$_),@x};for(
+$t=$s=0;$t-$s<$x[0];&$e(++$t)or$s=$t){}$s
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