TL;DR

Here we are with TASK #1 from The Weekly Challenge #126. Enjoy!

The challenge

You are given a positive integer $N. Write a script to print count of numbers from 1 to $N that donâ€™t contain digit 1.

Example

Input: $N = 15 Output: 8 There are 8 numbers between 1 and 15 that don't contain digit 1. 2, 3, 4, 5, 6, 7, 8, 9. Input:$N = 25
Output: 13

There are 13 numbers between 1 and 25 that don't contain digit 1.
2, 3, 4, 5, 6, 7, 8, 9, 20, 22, 23, 24, 25.


The questions

One question I would ask is whether there is a reasonable limit for the input number $N. I mean, this is extremely simple to break with a brute-force approach, but it becomes unfeasable with large number. On the same tune there would be the question about how many times should this be called in a speciic time slot. Nit-picking a bit, I guess that we have to count the integers between the two input numbers. I would double check that $N sould be indeed included, although the second example makes it pretty clear that it is indeed.

Last, I would also double check that weâ€™re talking base 10 here. Were it base 2â€¦ the answer would be trivial ðŸ¤“

The solution

The basic idea to address this puzzle is the observation that we can proceed in chunks. As an example, letâ€™s take 5314 as an example input.

It is greater than or equal to 5000, so for sure it will contain all matching number between 1 and 4999, plus those between 5000 and 5314.

Calculating the first slot can be simplified by observing that weâ€™re dealing with all the numbers whose first digit ranges from 0 to 4 and the rest of three digits range from 0 to 9. This also includes 0000 of course, but we can just subtract 1 from the result.

Now:

• the first digit must not contain 1, so out of the 4 in our example, we have to ignore that and be left with 0, 2, 3, and 4. That isâ€¦ 4 numbers. Where does the 4 come from, anyway? It is the first digit $F$ of our input number, less 1, so there we have it;
• the other digits can range from 0 to 9, skipping 1. So each slot can only contain 9 possible candidates, all independent of one another. If there are 3 slot as in our case, we have a total of $9^3$ allowed arrangement. In general, if the input number $N has$k$digits, this part will have$9^k$possible arrangements. All in all, then, we end up with$F \cdot 9^k$possible allowed arrangements. At this point, we can chop off the first digit 5 and do the same with the remaining part 314. Super-easyâ€¦ right?!? Wellâ€¦ almost. This is the general gist, but there are a ton of shady corners and special cases to consider. As an example, as soon as we arrive to a residual number starting with 1, it has no use to move further the basic calculation because all numbers will be prefixed with that 1. Considering that there are so many poorly lit corners, I decided to start with a brute-force baseline calculator, just to be able and double check my smarter solutions. I started with Raku: sub count-like-no-one-bf (Int:D$N where * > 0) {
(2 .. $N).grep({! /1/}).elems }  I donâ€™t know if itâ€™s idiomatic, but it works and itâ€™s readable. We start with the list of candidates (starting from 2 because 1 is out of the game, right?), remove any candidate with at least one 1 and count how many items weâ€™re left with. Theoretically speaking we would be done here, at least for inputs say below 1000 (arbitrary) and few calls. Practically speaking, though, itâ€™s time to move on. For technical reasons I switched to Perl, again starting with the brute force approach to have a baseline: sub count_like_no_ones_bf ($N) { scalar grep {! m{1}mxs} 2 .. $N }  The description of the algorithm made me think about a recursive implementation at first thought, then I realized that chopping off the first digit was actually very easy to accomplish with a loop, so I ended up with this: sub count_like_no_ones ($N) {
my $count = 0; my @digits = split m{}mxs,$N;
while (@digits) {
my $first = shift @digits; if (@digits) { # more to go after, use chunking my$factor = $first > 1 ?$first - 1 : $first;$count += $factor * 9 ** @digits; } else { # last digit, count all including 0$count += $first > 1 ?$first : 1;
}
last if ($first == 1); } # we took into account sequence of all 0, so we remove it return$count - 1;
}


I will never admit that it took me way more than I anticipated. Donâ€™t even ask please.

Anyway, it works although itâ€™s not as elegant as I would have liked. All the special cases are there, but they behave differently for the last digit and this is somehow itchy. Anyway.

At this point I wondered how the recursive implementation would look like. So, of course, I coded it:

sub count_like_no_ones_r ($N) { return($N > 1 ? $N - 1 : 0) if$N < 10;
my $first = substr$N, 0, 1, '';
my $factor =$first > 1 ? $first - 1 :$first;
my $count =$factor * 9 ** length($N);$count += 1 + count_like_no_ones_r($N) if$first != 1;
return $count - 1; }  This is probably slightly more intuitive. Making a difference between the general case and the one with one digit only is much simpler (first line in the sub). There is one little catch where we have to add 1 to the result of the recursive call, but itâ€™s needed to have the right value eventually so there it is. I think that the Benchmark is interesting in this case. I first started with computing all numbers from 1 to 9999: use Benchmark 'cmpthese'; ... my @inputs = 0 .. 9999; cmpthese(-5, { recursive => sub { count_like_no_ones_r($_) for @inputs },
iterative => sub { count_like_no_ones($_) for @inputs }, } );  and, surprise surprise, the recursive function worked better:  Rate iterative recursive iterative 35.4/s -- -8% recursive 38.4/s 8% --  Time and again Iâ€™m baffled by the fact that the recursive function works faster. Except that it does not scale as well as the iterative one, which can be seen by adding a few 9 before: my @inputs = 99999999990000 .. 99999999999999; cmpthese(-5, { recursive => sub { count_like_no_ones_r($_) for @inputs },
iterative => sub { count_like_no_ones($_) for @inputs }, } );  Running it now yields:  Rate recursive iterative recursive 8.67/s -- -22% iterative 11.0/s 27% --  which is what I was expecting to be honest. Last, I returned to Raku, only porting the iterative version in what is basically a straight translation with very few changes: sub count-like-no-one (Int:D$N where * > 0) {
my $count = 0; my @digits =$N.comb;
while (@digits) {
my $first = @digits.shift; if (@digits) { my$factor = $first > 1 ??$first - 1 !! $first;$count += $factor * 9 ** @digits; } else {$count += $first > 1 ??$first !! 1;
}
last if $first == 1; } return$count - 1;
}


And with thisâ€¦ I guess itâ€™s all for this puzzle! I hope you enjoyed the ride and that will accept a recommendationâ€¦ to stay safe!