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moritz's post, or rather his discussion of it on the ChatterBox, made me think, and it inspired me to come up with a related yet different solution.

My solution is based on the idea of a step function, often spelled "�" in math, and adding them together (function of x, edge at t):

```    �(x, t) = 0 if x < t
= 1 if x > t
= ?? if x == t (often 1/2 in continuous math functions)
```
As this is about discrete values, I'd rather make that
```    �(x, t) = 0 if x < t
= 1 if x > t
= 1 if x == t (for discrete input)
```
That way, the added value for a range, say [20, 50], can be
```    �(x, 20) - �(x, 51)
```
So, just as with moritz's solution, you can store the value for a range as 2 tuples for each edge:
```%step = ( 20 => 1, 50+1 => -1 );
You can imagine this to represent a number of Dirac pulses; the combination of step functions is the integral of this set of pulses, or, because we're in the discrete case, the sum of amplitudes of every pulse on its left.

For each range you add, in the same hash, you can then increment the value for the lower edge, and decrement for the upper edge + 1.

```my %step;
foreach(@range) {
\$step{ \$_-> }++;
\$step{ \$_->+1 }--;
}
Now, the next step is to convert this data structure to the final outcome. If you need a complete data structure for the whole spectrum, you'll have to loop through the full range once, on each edge, adding the value for \$step{\$edge} to the previous accumulated value and assigning that to the range:
```my @edge = sort { \$a <=> \$b } keys %step;
my \$step = 0;
my \$x = shift @edge;
for(my \$i = \$x; defined \$x; \$i++) {
if(\$i == \$x) {
\$step += \$step{\$x};
\$x = shift @edge;
}
\$arr[\$i] += \$step;
}
Tested with
```@arr = (0) x 120;
@range = ([20, 50], [30, 100]);
The end result of print "@arr\n"; is:
```0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2
+2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
+ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0
+0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

In reply to Re: Optimizing a double loop by bart
in thread Optimizing a double loop by roibrodo

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