TL;DR

Where we solve Perl Weekly Challenge #079 task #2.

# The challenge

You are given an array of positive numbers @N.

Write a script to represent it as Histogram Chart and find out how much water it can trap.

# The questions

I think that there’s a bulk of questions that we should always ask:

• are there limits on the size of the array?

# The solution

One possible simplistic approach is to ask: how much water do we find at each location? This is easily determined: let’s look at the maximum height on the left (including where we are standing), then let’s look at the maximum height on the right (again, inclucing where we are standing). The minimum of these two maxima will be the height of the water; subtract the height of the spot and we have the answer. Repeat for all locations… and we’re done.

This can be translated into code in a quite straightforward way:

 1 sub trapped_rain_water_dumb (@N) {
2    my $retval = 0; 3 for my$i (1 .. $#N - 1) { 4 my$max_left  = max(@N[0 .. $i]); 5 my$max_right = max(@N[$i ..$#N]);
6       my $min_max = min($max_left, $max_right); 7$retval += $min_max -$N[$i]; 8 } 9 return$retval;
10 }


I’m calling this implementation dumb because it’s quite inefficient. Calculating the maximum value for all spots in both direction time and again is… suboptimal. We are iterating over the array (external loop starting in line 3) and then we are iterating again on the array (internal loop, divided between lines 4 and 5). Overall… not an encouraging complexity ($O(N^2)$). Anyway… a good benchmark.

So how can we make it better? Let’s reverse the perspective: we can sweep the array from left to right and set the maximum value found for each location as coming from the left. Then we do a similar sweep from the right, and update the maximum value taking the minor of the two. This means that we can do just two sweeps in total and have an asyntotic complexity of $O(N)$… yay!

Which leads us to the following code:

 1 sub trapped_rain_water (@N) {
2    my $max = 0; 3 my @maxes; 4 5 # first pass, left to right 6 for my$v (@N) {
7       $max = max($max, $v); 8 push @maxes,$max;
9    }
10
11    # second pass, right to left
12    my $retval = 0; 13$max = 0;
14    for my $v (reverse @N) { 15$max = max($max,$v);
16       my $w = min($max, shift @maxes);
17       $retval +=$w - $v; 18 } 19 return$retval;
20 }


In the first pass (lines 5 through 9) we set the maximum value coming from the left, saving it into @maxes.

Then, we do a second pass from right to left, calculate the maximum value “from the right” for each spot and then take the minimum of the two maxima (line 16).

Now, the difference between this “maximum minimum” and the height at the spot is how much water is held in place, so we can add thsi value to the total returned.

I hope you enjoyed it!